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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`1/(c o s e ctheta-cottheta)-1/(sintheta)=1/(sintheta)-1/(c o s e ctheta+cottheta)` |
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Answer» We have `LHS=(1)/(("cosec" theta - cot theta))-(1)/(sin theta) ` `=(1)/(("cosec" theta - cot theta)) xx (("cosec" theta + cot theta))/(("cosec" theta + cot theta)) -(1)/(sin theta) ` ` =(("cosec" theta + cot theta))/(("cosec"^(2)theta - cot^(2)theta)) - "cosec" theta " "[because (1)/(sin theta)="cosec" theta] ` ` = ("cosec" theta + cot theta)- "cosec" theta " "[because "cosec"^(2)theta - cot^(2)theta =1 ] ` ` = cot theta. ` `RHS=(1)/(sin theta)-(1)/(("cosec" theta + cot theta))` `="cosec" theta - (1)/(("cosec" theta + cot theta))xx(("cosec" theta - cot theta))/(("cosec" theta - cot theta))` `= "cosec" theta - (("cosec" theta - cot theta))/(("cosec"^(2)theta - cot^(2)theta)) ` `="cosec" theta-("cosec" theta - cot theta) " " [ because "cosec"^(2)theta-cot^(2)theta=1] ` `= cot theta.` ` therefore LHS = RHS. ` |
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| 2. |
If ` x=r sin alpha cos beta, y= r sin alpha sin beta and z= r cos alpha , ` prove that ` x^(2)+y^(2) + z^(2) = r^(2). ` |
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Answer» We have ` x^(2) + y^(2) + z^(2) = r^(2) sin^(2) alpha cos^(2) beta +r^(2) sin^(2) alpha sin^(2) beta + r^(2) cos^(2) alpha ` ` = r^(2)sin^(2) alpha (cos^(2) beta + sin^(2) beta) + r^(2) cos^(2) alpha ` `= r^(2) sin^(2) alpha + r^(2) cos^(2) alpha " " [because cos^(2) beta + sin^(2) beta =1] ` `= r^(2) (sin^(2) alpha + cos^(2) alpha )= r^(2) " " [because sin^(2) alpha + cos^(2) alpha =1]. ` Hence, ` (x^(2) + y^(2) + z^(2))= r^(2) . ` |
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| 3. |
If `cos e ctheta-sintheta=ma n dsectheta-costheta=n ,`prove that `(m^(2n))^(2/3)+(m m^2)^(2/3)=1` |
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Answer» We have ` m^(2)n = ("cosec" theta - sin theta)^(2)* (sec theta - cos theta) ` ` =((1)/(sin theta) - sin theta)^(2) * ((1)/(cos theta) - cos theta) ` `= ((1- sin^(2) theta)^(2) )/(sin^(2)theta) * ((1- cos^(2)theta))/(cos theta) = (cos^(4)theta)/(sin^(2)theta) xx (sin^(2) theta)/(cos theta) = cos^(3) theta ` ` therefore (m^(2)n)^(1//3) = cos theta . " "...(i) ` Again, `mn^(2) = ("cosec" theta - sin theta)* (sec theta - cos theta)^(2) ` `= ((1)/(sin theta)-sin theta) * ((1)/(cos theta) - cos theta)^(2) ` `= ((1- sin^(2) theta))/(sin theta) * ((1- cos^(2)theta)^(2))/(cos^(2) theta) ` `= ((cos^(2) theta)/(sin theta) xx (sin^(4) theta)/(cos^(2) theta)) = sin^(3) theta ` ` therefore (mn^(2))^(1//3) = sin theta . " "...(ii) ` On squaring (i) and (ii) and adding the results, we get ` (m^(2)n)^(2//3) + (mn^(2))^(2//3) = 1 " "[because cos^(2) theta + sin^(2) theta =1]. ` Hence,`(m^(2)n)^(2//3) + (mn^(2))^(2//3) =1. ` |
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| 4. |
If `(3 sin theta + 5 cos theta ) = 5, ` prove that ` (5 sin theta - 3 cos theta ) = pm 3. ` |
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Answer» We have `(3 sin theta + 5 cos theta)^(2) + (5 sin theta - 3 cos theta)^(2) ` ` = 9 (sin^(2) theta + cos^(2) theta ) + 25 ( sin^(2) theta + cos^(2) theta) ` ` = (9+ 25)= 34.` ` therefore (3 sin theta + 5 cos theta)^(2) + ( 5 sin theta - 3 cos theta)^(2) =34 ` ` rArr 5^(2) + (5 sin theta - 3 cos theta)^(2) = 34 " " [ because 3 sin theta + 5 cos theta = 5 ] ` ` rArr (5 sin theta - 3 cos theta) = pm 3 " " `[ taking square root on each side] Hence, `(5 sin theta - 3 cos theta) = pm 3. ` |
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| 5. |
If a `costheta-bsintheta=c ,`prove that a `sintheta+bcostheta=+-sqrt(a^2+b^2-c^2)` |
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Answer» Given, ` a cos theta - b sin theta = c. " "...(i) ` Now, ` (a cos theta - b sin theta)^(2) + (a sin theta + b cos theta)^(2) ` `=a^(2) (cos^(2) theta + sin^(2) theta) + b^(2) (sin^(2) theta + cos ^(2) theta)= (a^(2) +b^(2)) .` Thus, ` (a cos theta - b sin theta)^(2) + (a sin theta + b cos theta)^(2) = (a^(2) + b^(2)) ` `rArr c^(2) + (a sin theta + b cos theta)^(2) = (a^(2) + b^(2)) ` `rArr (a sin theta + b cos theta)^(2) = (a^(2) + b^(2) - c^(2))` ` rArr (a sin theta + b cos theta ) = pm sqrt(a^(2) + b^(2) - c^(2)). ` Hence, `(a sin theta + b cos theta) = pm sqrt(a^(2) + b^(2) - c^(2)). ` |
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| 6. |
Choose the correct answer in each of the following questions: `(sec 30^(@))/("cosec"60^(@))=?`A. `(2)/(sqrt(3))`B. `(sqrt(3))/(2)`C. `sqrt(3)`D. 1 |
| Answer» Correct Answer - D | |
| 7. |
`cos1^@xxcos2^@xxcos3^@xx...xxcos180^@=?`A. -1B. 1C. 0D. `(1)/(2)` |
| Answer» Correct Answer - C | |
| 8. |
Choose the correct answer in each of the following questions: `sin47^(@)cos43^(@)+cos47^(@)sin43^(@)=?`A. `sin4^(@)`B. `cos4^(@)`C. 1D. 0 |
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Answer» Correct Answer - C Given expression `=sin47^(@)cos(90^(@)-47^(@))+cos47^(@)sin(90^(@)-47^(@))` `=(sin^(2)47^(@)+cos^(2)47^(@))=1.` |
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| 9. |
Prove that `(sin theta + "cosec" theta)^(2)+(cos theta + sec theta)^(2)=(7+ tan^(2) theta + cot^(2) theta).` |
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Answer» We have LHS `= (sin theta + "cosec" theta)^(2)+(cos theta + sec theta)^2 ` ` =(sin^(2)theta + "cosec"^(2)theta +2 sin theta "cosec" theta)+(cos^(2)theta + sec^(2)theta + 2 cos theta sec theta )` ` = (sin^(2) theta + "cosec"^(2)theta +2)+(cos^(2)theta + sec^(2)theta +2) ` ` [ because sin theta "cosec"theta =1 and cos theta sec theta=1] ` ` = (sin^(2)theta + cos^(2)theta)+4+("cosec"^(2) theta + sec^(2) theta)` ` = 1+4+(1+cot^(2) theta)+(1+tan^(2)theta) ` ` [ because sin^(2) theta + cos^(2)theta =1, "cosec"^(2)theta =1+cot^(2) theta and sec^(2)theta=1+tan^(2)theta] ` ` =(7+ tan^(2)theta + cot^(2) theta )= RHS. ` `therefore LHS = RHS. ` |
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| 10. |
Prove that `(i) (1- sin theta)/(1+ sin theta)=(sec theta - tan theta )^(2) ` `(ii) ((1+cos theta))/((1- cos theta))= ("cosec" theta + cot theta)^(2)` |
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Answer» We have `(i) RHS = (sec theta - tan theta)^(2) ` ` =((1)/(cos theta)-(sin theta)/(cos theta))^(2)=((1-sin theta)/(cos theta))^(2)=((1- sin theta)^(2))/(cos^(2)theta)` ` = ((1-sin theta)^(2))/((1-sin^(2)theta))=((1- sin theta)(1-sin theta))/((1+ sin theta)(1-sin theta))` `=((1-sin theta))/((1+sin theta))= LHS. ` ` therefore RHS = LHS. ` `(ii) RHS= ("cosec" theta + cot theta)^(2) =((1)/(sin theta)+(cos theta)/(sin theta))^(2) ` ` =((1+ cos theta)/(sin theta))^(2) = ((1+ cos theta)^(2))/(sin^(2)theta) =((1+ cos theta)(1+ cos theta))/(1- cos^(2)theta)` ` =((1+ cos theta)(1+ cos theta))/((1+ cos theta)(1- cos theta))=((1+ cos theta))/((1- cos theta))= LHS. ` `therefore LHS = RHS. ` |
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| 11. |
`(1-cos^2 theta) cosec^2 theta=1` |
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Answer» Correct Answer - 1 `(1-cos^(2)theta)"cosec"^(2)theta=sin^(2)theta*"cosec"^(2)theta=1.` |
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| 12. |
Evaluate : `(2sin^2 63^@+1+2sin^2 27^@)/(3 cos ^2 17^@-2+3cos^2 73^@)`A. `(3)/(2)`B. `(2)/(3)`C. 2D. 3 |
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Answer» Correct Answer - D Given expression `=(2(sin^(2)63^(@)+sin^(2)27^(@))+1)/(3(cos^(2)17^(@)+cos^(2)73^(@))-2)=(2(sin^(2)63^(@)+cos^(2)63^(@))+1)/(3(cos^(2)17+sin^(2)17^(@))-2)` `=((2xx1)+1)/((3xx1)-2)=3. " " |(because sin27^(@)=sin(90^(@)-63^(@))=cos63^(@)),(cos73^(@)=cos(90^(@)-17^(@))=sin17^(@))|` |
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| 13. |
Write the value of `(1+cot^(2)theta)sin^(2)theta.` |
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Answer» Correct Answer - 1 `(1+cot^(2)theta)sin^(2)theta="cosec"^(2)thetasin^(2)theta=1.` |
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| 14. |
Write thevalue of `cot^2theta-1/(sin^2theta)`. |
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Answer» Correct Answer - -1 `(cot^(2)theta-(1)/(sin^(2)theta))=(cot^(2)theta-"cosec"^(2)theta)=cot^(2)theta-(1+cot^(2)theta)=-1.` |
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| 15. |
If `2sin 2 theta =sqrt(3) " then " theta=?`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - A `2sin2theta=sqrt(3)rArr sin2 theta=(sqrt(3))/(2)=sin60^(@)rArr 2 theta=60^(@)rArrtheta=30^(@).` |
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| 16. |
What is thevalue of `sin^2theta+1/(1+tan^2theta)`? |
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Answer» Correct Answer - 1 `(sin^(2)theta+(1)/(1+tan^(2)theta))=(sin^(2)theta+(1)/(sec^(2)theta))=(sin^(2)theta+cos^(2)theta)=1.` |
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| 17. |
If `sintheta+sin^2theta=1` Prove that `cos^2theta+cos^4theta=1` |
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Answer» ` sin theta + sin^(2)theta =1 ` `rArr sin theta= 1- sin^(2)theta ` `rArr sin theta = cos^(2)theta " " [because 1- sin^(2)theta = cos^(2)theta] ` `rArr sin^(2)theta = cos^(4)theta ` `rArr 1- cos^(2)theta =cos^(4)theta " " [because sin^(2)theta = 1- cos^(2)theta] ` `rArr cos^(2)theta +cos^(4) theta =1. ` Hence, ` cos^(2)theta + cos^(4)theta =1. ` |
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| 18. |
`(cosec theta)/(cosec theta-1)+(cosec theta)/(cosec theta+1)=2sec^2theta` |
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Answer» We have `LHS= ("cosec" theta)/(("cosec" theta -1))+ ("cosec" theta)/(("cosec" theta +1)) ` ` = ("cosec" theta ("cosec" theta +1)+ "cosec" theta("cosec" theta -1))/(("cosec"^(2)theta -1)) ` ` = (2 "cosec"^(2)theta)/((1+ cot^(2)theta -1)) " " [ because "cosec"^(2)theta=1+cot^(2)theta] ` `= (2"cosec"^(2)theta)/(cot^(2)theta)=2 "cosec"^(2)theta tan^(2)theta ` `= 2 xx (1)/(sin^(2)theta) xx (sin^(2)theta)/(cos^(2)theta)=(2)/(cos^(2)theta) = 2 sec^(2) theta = RHS. ` |
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| 19. |
Choose the correct answer in each of the following questions: If `2cos 3 theta=1 " then " theta=?`A. `10^(@)`B. `15^(@)`C. `20^(@)`D. `30^(@)` |
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Answer» Correct Answer - C `2cos3theta=1rArr cos3theta=(1)/(2)=cos60^(@)rArr3theta=60^(@)rArr theta=20^(@).` |
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| 20. |
If `tanx= 3cotx " then " x=?`A. `45^(@)`B. `60^(@)`C. `30^(@)`D. `15^(@)` |
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Answer» Correct Answer - B `tanx=3cotxrArr tan^(2)x=3rArr tanx=sqrt(3)=tan60^(@)rArrx=60^(@).` |
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| 21. |
Find the value of `[(sin^2 2 2^(@)+sin^2 6 8^(@))/(cos^2 2 2^(@)+cos^2 6 8^(@))+sin^2 6 3^(@)+cos6 3^(@)sin2 7^(@)]` |
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Answer» Correct Answer - C Given expression `=(sin^(2)22^(@)+sin^(2)(90^(@)-22^(@)))/(cos^(2)22^(@)+cos^(2)(90^(@)-22^(@)))+sin^(2)63^(@)+cos63^(@)sin(90^(@)-63^(@))` `=(sin^(2)22^(@)+cos^(2)22^(@))/(cos^(2)22^(@)+sin^(2)22^(@))+sin^(2)63^(@)+cos^(2)63^(@)=(1)/(1)+1=1+1=2.` |
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| 22. |
Choose the correct answer in each of the following questions: If `sqrt(3) tan 2theta-3=0 " then " theta=?`A. `15^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - B `sqrt(3)tan 2theta=3rArr tan 2theta=(3)/(sqrt(3))=sqrt(3)=tan60^(@)rArr 2theta=60^(@)rArrtheta=30^(@)`. |
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| 23. |
Prove each of the following identities : `(1+ tan theta + cot theta )( sin theta - cos theta ) =((sec theta)/("cosec"^(2) theta ) - ("cosec"theta)/(sec^(2)theta)) ` |
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Answer» `LHS=(1+(sintheta)/(costheta)+(costheta)/(sintheta))(sintheta -costheta)=((sinthetacostheta+sin^(2)theta+cos^(2)theta)(sintheta-costheta))/(sinthetacostheta)` `=((sin^(3)theta -cos^(3)theta))/(sinthetacostheta)=((sin^(3)theta)/(sinthetacostheta)-(cos^(3)theta)/(sinthetacostheta))=((sin^(2)theta)/(costheta)-(cos^(2)theta)/(sintheta))` `=((sectheta)/("cosec"^(2)theta)-("cosec"^(2)theta)/(sec^(2)theta))=RHS.` |
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| 24. |
Prove the following identity: `(1/(sec^2theta-cos^2theta)+1/(cos e c^2theta-sin^2theta))sin^2thetacos^2theta=(1-sin^2thetacos^2theta)/(2+sin^2cos^2theta)` |
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Answer» `LHS={(1)/(((1)/(cos^(2)theta)-cos^(2)theta))+(1)/(((1)/(sin^(2)theta)-sin^(2)theta))}(sin^(2)thetacos^(2)theta)` `={(cos^(2)theta)/((1-cos^(4)theta))+(sin^(2)theta)/((1-sin^(4)theta))}sin^(2)theta cos^(2)theta` `={(cos^(2)theta)/(sin^(2)theta(1+cos^(2)theta))+(sin^(2)theta)/(cos^(2)theta(1+sin^(2)theta))}sin^(2)thetacos^(2)theta` `=(cos^(4)theta(1+sin^(2)theta)+sin^(4)theta(1+cos^(2)theta))/((1+cos^(2)theta)(1+sin^(2)theta))` `=(cos^(4)theta+sin^(4)theta+cos^(2)thetasin^(2)theta(cos^(2)theta+sin^(2)theta))/(1+(sin^(2)theta+cos^(2)theta)+sin^(2)thetacos^(2)theta)` `=((cos^(2)theta+sin^(2)theta)^(2)-sin^(2)thetacos^(2)theta)/(2+sin^(2)theta cos^(2)theta)=(1-sin^(2)thetacos^(2)theta)/(2+sin^(2)theta cos^(2)theta)` |
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| 25. |
Prove each of the following identities : `(cot^(2) theta (sec theta -1))/((1+ sin theta))+ (sec^(2)theta (sintheta -1))/((1+ sec theta)) =0 ` |
| Answer» `LHS =(cot^(2)theta(sec^(2)theta-1)+sec^(2)theta (sin^(2)theta-1))/((1+sintheta)(1+sec theta))=((cot^(2)thetatan^(2)theta)-(sec^(2)thetacos^(2)theta))/((1+sintheta)(1+sec theta))=0.` | |
| 26. |
If `cottheta=1/(sqrt(3))`, show that`(1-cos^2theta)/(2-sin^2theta)=3/5` |
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Answer» Correct Answer - `(3)/(5)` `((1-cos^(2)theta))/((2-sin^(2)theta))=(sin^(2)theta)/((2-sin^(2)theta))=((3)/(4))/((2-(3)/(4)))" "["cosec"^(2)theta=(1+cot^(2)theta)=(1+(1)/(3))=(4)/(3) rArr sin^(2)theta=(3)/(4)]` `=((3)/(4)xx(4)/(5))=(3)/(5).` |
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| 27. |
If ` 3 cot theta =4, " write the value of " ((2costheta+ sintheta ))/((4costheta- sintheta)).` |
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Answer» Correct Answer - `(11)/(13)` `((2costheta+sintheta))/((4costheta+sintheta))=((2cottheta +1))/((4cottheta-1))" "["dividing num. and denom. by "sintheta]` `=((2xx(4)/(3)+1)/(4xx(4)/(3)-1))=(11)/(13).` |
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| 28. |
If `5 tan theta =4, " write the value of " ((costheta- sintheta ))/((costheta+ sintheta)).` |
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Answer» Correct Answer - `(1)/(9)` `((costheta-sintheta))/((costheta+sintheta ))=((1-tantheta))/((1+tantheta))" "["dividing num. denom. by cos"theta]` `=((1-(4)/(5)))/((1+(4)/(5)))=(1)/(9) " "[because tantheta=(4)/(5)].` |
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| 29. |
If `costheta+sintheta`=`sqrt2sintheta` then `sintheta-costheta` is |
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Answer» `(costheta+sintheta)^(2)=2sin^(2)theta rArr cos^(2)theta+sin^(2)theta+2costhetasintheta=2 sin^(2)theta` `rArr sin^(2)theta-2costhetasintheta= cos^(2)theta rArr sin^(2)theta - 2costhetasintheta+cos^(2)theta=2cos^(2)theta.` `rArr (sintheta-costheta)^(2)=2cos^(2)theta rArr (sintheta- costheta )=sqrt(2) costheta.` |
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| 30. |
The value of `tan 10^(@)* tan20^(@)*tan45^(@)*tan70^(@)*tan80^(@)`=_________ |
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Answer» Correct Answer - 1 Given expression `=(tan 10^(@)tan80^(@))(tan20^(@)tan70^(@))` `={tan 10^(@) tan(90^(@)-10^(@))}{tan20^(@)*tan(90^(@)-20^(@))}` `=(tan10^(@) cot10^(@))(tan20^(@)cot 20^(@))=(1xx1)=1.` |
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| 31. |
Prove each of the following identities : `( cos^(2) theta)/((1- tan theta)) + (sin^(3) theta)/((sin theta - cos theta)) = (1+ sin theta cos theta) ` |
| Answer» Write ` tan theta = (sin theta )/(cos theta ) ` and simplify . | |
| 32. |
Prove each of the following identities : ` (cos theta) /((1- tan theta)) - (sin^(2) theta)/((cos theta - sin theta)) = (cos theta + sin theta) ` |
| Answer» Write ` tan theta = (sin theta )/(cos theta ) ` and simplify . | |
| 33. |
Prove each of the following identities : ` (tan^(2) theta)/((1+ tan^(2) theta)) + (cot^(2) theta)/((1+ cot^(2) theta)) =1 ` |
| Answer» `LHS=(tan^(2)theta)/(sec^(2)theta)+(cot^(2)theta)/("cosec"^(2)theta)=((sin^(2)theta)/(cos^(2)theta)xxcos^(2)theta)+((cos^(2)theta)/(sin^(2)theta)xxsin^(2)theta)=(sin^(2)theta+cos^(2)theta)=1.` | |
| 34. |
Prove each of the following identities : `(1+ tan^(2) theta)(1+ cot^(2) theta)=(1)/((sin^(2) theta- sin^(4) theta)) ` |
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Answer» LHS ` = sec^(2) theta * "cosec"^(2) theta = (1)/(cos^(2) theta sin^(2) theta ) ` ` = (1)/(sin^(2)theta (1-sin^(2) theta) ) = (1)/((sin^(2) theta - sin^(4) theta )) .` |
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| 35. |
Prove each of the following identities : `(tan theta)/((1+ tan^(2) theta)^(2)) + (cot theta)/((1+ cot^(2) theta)^(2)) = sin theta cos theta ` |
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Answer» LHS ` = (tan theta )/(sec^(4) theta ) + (cot theta )/("cosec"^(4) theta ) = ((sin theta)/(cos theta) xx cos^(4) theta ) + ((cos theta)/(sin theta) xx sin^(4) theta) ` ` = (sin theta cos^(3) theta + sin^(3) theta cos theta ) = sin theta cos theta ( cos^(2) theta + sin^(2) theta ) = sin theta cos theta . ` |
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| 36. |
Prove each of the following identities : ` (i) sin^(6) theta + cos^(6)theta = 1- 3 sin^(2) theta cos^(2) theta ` `(ii) sin^(2)theta + cos^(4) theta = cos^(2) theta + sin^(4) theta ` `(iii) "cosec"^(4) theta - "cosec"^(2) theta = cot^(4) theta + cot^(2) theta ` |
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Answer» (i) `sin^(6)theta+cos^(6)theta=(sin^(2)theta)^(3)+(cos^(2)theta)^(3)` `=(sin^(2)theta+cos^(2)theta)(sin^(4)theta+cos^(4)theta-sin^(2)thetacos^(2)theta)` `=(sin^(2)theta+cos^(2)theta)^(2)-3 sin^(2)thetacos^(2)theta=(1-3sin^(2)thetacos^(2)theta).` (ii) `(cos^(4)theta-sin^(4)theta)=(cos^(2)theta-sin^(2)theta)(cos^(2)theta+sin^(2)theta)` `rArr (cos^(4)theta-sin^(4)theta)=(cos^(2)theta-sin^(2)theta)` `rArr (sin^(2)theta+cos^(4)theta)=(cos^(2)theta+sin^(4)theta).` (iii) `("cosec"^(4)theta-cot^(4)theta)=("cosec"^(2)theta+cot^(2)theta)("cosec"^(2)theta-cot^(2)theta) ` `rArr ("cosec"^(4)theta-cot^(4)theta)=("cosec"^(2)theta +cot^(2)theta) " " [ because "cosec"^(2)theta-cot^(2)theta=1]` `rArr ("cosec"^(4)theta-"cosec"^(2)theta)=(cot^(4)theta+cot^(2)theta).` |
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| 37. |
Prove each of the following identities : `(i) (1+ cos theta + sin theta)/(1+ cos theta - sin theta) =(1+ sin theta)/(cos theta) ` `(ii) (sin theta + 1- cos theta)/(cos theta - 1 + sin theta) = (1+ sin theta)/(cos theta) ` |
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Answer» (i) On dividing num. and denom. by `cos theta`, we get `LHS=((sec theta+tan theta)+1)/((sec theta+1-tan theta))=((sectheta+tan theta)+(sec^(2)theta-tan^(2)theta))/((sec theta+1-tan theta))` `=((sec theta+tan theta)(1+sectheta-tan theta))/((1+sectheta-tan theta))=(sec theta+tan theta)=((1)/(cos theta)+(sintheta)/(cos theta)).` (ii) On dividing num. and denom. by `cos theta`, we get `LHS =(tantheta+sectheta-1)/(1-sectheta+tantheta)=((sectheta+tantheta)-(sec^(2)theta-tan^(2)theta))/((tantheta-sec theta+1))` `=((sectheta+tantheta)[1-(sectheta-tantheta)])/((tan theta-sectheta+1))=(sec theta+tan theta).` |
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| 38. |
Prove each of the following identities : ` (1+ cos theta - sin^(2) theta)/(sin theta (1+ cos theta)) = cot theta ` |
| Answer» Write `sin^(2)theta =(1-cos^(2)theta).` | |
| 39. |
Prove each of the following identities : `(i) (sec theta -1)/(sec theta +1) =(sin^(2) theta)/((1+ cos theta)^(2) ) ` ` (ii) (sec theta - tan theta)/(sec theta + tan theta) = (cos^(2) theta)/((1+ sintheta)^(2)) ` |
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Answer» (i) `LHS =(((1)/(costheta)-1))/(((1)/(costheta)+1))=((1-costheta))/((1+cos theta))xx((1+costheta))/((1+cos theta))=((1-cos^(2)theta))/((1+cos theta)^(2))=(sin^(2)theta)/((1+cos theta)^(2)).` (ii) `LHS =(((1)/(costheta)-(sin theta)/(costheta)))/(((1)/(costheta)+(sin theta)/(costheta)))=((1-sin theta))/((1+sin theta))xx((1+sintheta))/((1+sin theta))=((1-sin^(2)theta))/((1+sin theta)^(2))=(cos^(2)theta)/((1+sin theta)^(2)).` |
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| 40. |
Prove each of the following identities : `(cos^(3) theta + sin^(3) theta)/(cos theta + sin theta ) + (cos^(3) theta - sin^(3) theta)/(cos theta - sin theta ) = 2 ` |
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Answer» `(cos^(3)theta+sin^(3)theta)=(costheta+sintheta)(cos^(2)theta+sin^(2)theta-costhetasintheta).` And, `(cos^(3)theta-sin^(3)theta)=(cos theta-sintheta) (cos^(2)theta+sin^(2)theta+cos theta sin theta).` |
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| 41. |
Prove each of the following identities : `(i) sqrt((1+ sin theta)/(1- sin theta)) = ( sec theta + tan theta) ` ` (ii) sqrt((1- cos theta)/(1+ cos theta))= ("cosec" theta - cot theta) ` ` (iii) sqrt((1+ cos theta)/(1- cos theta))+ sqrt((1- cos theta)/(1+ cos theta)) = 2 "cosec" theta ` |
| Answer» (i) `LHS=(sqrt(1+sintheta))/(sqrt(1-sintheta))xx(sqrt(1+sintheta))/(sqrt(1+sintheta))=((1+sin theta))/(sqrt(1-sin^(2)theta))=((1+sin theta))/(cos theta)=((1)/(cos theta)+(sintheta)/(costheta))=(sec theta+tan theta).` | |
| 42. |
If ` sin theta =x, " write the value of " cot theta. ` |
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Answer» Correct Answer - `(sqrt(1-x^(2)))/(x)` `costheta=sqrt(1-sin^(2)theta)=sqrt(1-x^(2)).` `therefore cot theta=(costheta)/(sintheta)=(sqrt(1-x^(2)))/(x).` |
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| 43. |
If `sec theta =x, " write the value of " tan theta.` |
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Answer» Correct Answer - `sqrt(x^(2)-1)` `sec^(2)theta=(1+tan^(2)theta)rArrtan^(2)theta=(sec^(2)theta-1)` `rArr tantheta=sqrt(sec^(2)theta-1)=sqrt(x^(2)-1).` |
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| 44. |
`(tan 35^(@))/(cot 55^(@))+(cot 78^(@))/(tan 12^(@))=?` |
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Answer» Correct Answer - C Given expression `=(tan35^(@))/(cot(90^(@)-35^(@)))+(cot78^(@))/(tan(90^(@)-78^(@)))=(tan35^(@))/(tan35^(@))+(cot78^(@))/(cot78^(@))=(1+1)=2.` |
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| 45. |
If `cos theta=(2)/(3), " write the value of " (4+ 4 tan^(2)theta).` |
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Answer» Correct Answer - 9 `(4+4tan^(2)theta)=4(1+tan^(2)theta)=4sec^(2)theta=(4xx(9)/(4))=9" "[because sectheta=(3)/(2)].` |
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| 46. |
If `sin theta=(1)/(2), " write the value of " (3cot^(2)theta+3).` |
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Answer» Correct Answer - 12 `3 cot^(2)theta+3=3(cot^(2)theta+1)=3"cosec"^(2)theta=(3xx4)=12 " "[because "cosec"theta=2].` |
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| 47. |
Write the value of `(tan^(2)theta-sec^(2)theta)/(cot^(2)theta-"cosec"^(2)theta).` |
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Answer» Correct Answer - 1 Given expression `=((sec^(2)theta-tan^(2)theta))/(("cosec"^(2)theta-cot^(2)theta))=(1)/(1)=1.` |
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| 48. |
Prove that ` (cos theta)/((1-tan theta))+(sin theta)/((1-cot theta))=(cos theta+sin theta). ` |
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Answer» We have ` "LHS " =(cos theta)/((1- tan theta))+(sin theta)/((1-cot theta))=(cos theta)/((1-(sin theta)/(cos theta)))+(sin theta)/((1- (cos theta)/(sin theta)))` ` =(cos ^(2)theta)/((cos theta- sin theta))+(sin^(2) theta)/((sin theta - cos theta))` ` =(cos^(2) theta)/(cos theta - sin theta)-(sin^(2) theta)/(cos theta - sin theta)` `=(cos^(2)theta - sin^(2)theta)/((cos theta - sin theta))=((cos theta- sin theta)(cos theta +sin theta))/((cos theta- sin theta)) ` ` =(cos theta+sin theta)= " RHS. " ` ` therefore "LHS " = " RHS." ` |
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| 49. |
Write the value of ` 4 tan^(2)theta-(4)/(cos^(2)theta).` |
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Answer» Correct Answer - -4 Given expression `=4tan^(2)theta-4sec^(2)theta=4tan^(2)theta-4(1+tan^(2)theta)= -4.` |
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| 50. |
Write the value of ` "cosec"^(2)(90^(@)-theta)-tan^(2)theta.` |
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Answer» Correct Answer - 1 ` "cosec"^(2)(90-theta)-tan^(2)theta=sec^(2)theta-tan^(2)theta=1.` |
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