InterviewSolution
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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. |
If `sin theta=-(1)/(2)` and ` tan theta=(1)/sqrt(3)` then `theta` is equal to :-A. `30^(@)`B. `150^(@)`C. `210^(@)`D. none of these |
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Answer» Correct Answer - C Let us first out `theta` lying between 0 to `360^(@)` Since `sin theta=-(1)/(2)rArrtheta=210 ^(@)` or `330^(@)` and `tan theta=(1)/sqrt(3)rArr0=30^(@)` or `210^(@)` Hence, `theta=210^(@)` or `(7pi)/(6)` is the value satisfying both. |
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| 2. |
Find the value of `6(sin^6theta+cos^6theta)-9(sin^4theta+cos^4theta)+4` |
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Answer» Correct Answer - C `4[(sin^(2)theta+cos^(2)theta)^(3)-3sin^(2)thetacos^(2)theta(sin^(2)theta+cos^(2)theta)]-6[(sin^(2)theta+cos^(2)theta)^(2)-2sin^(2)thetacos^(2)theta)]` `=4[1-3sin^(2)thetacos^(2)theta]-[1-2sin^(2)thetacos^(2)theta]` `=4-12sin^(2)thetacos^(2)theta-6+12sin^(2)thetacos ^(2)theta=-2` |
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| 3. |
If `s intheta+sin^2theta1=1,`then prove that `cos^(12)theta+3cos^8theta+cos^6theta-1=0`Given that `s i lntheta=1-sin^2theta=1-sin^2theta=cos^2theta` |
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Answer» Given that `sin theta=1-sin^(2)theta=cos^(2)theta` LHS= `cos^(6)theta(ccos^(2)theta+1)^(3)-1=sin^(3)theta(1 +sintheta)^(3)-1=(sin theta+sin^(2)theta)^(3)-1=1-1=0` |
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| 4. |
P(x)=`(x-cos36^@)(x-cos84^@)(x-cos156^@)` then coefficient of `x^2` isA. `(3)/(2)`B. `-(3)/(2)`C. `-(3)/(4)`D. zero |
| Answer» Correct Answer - C | |
| 5. |
P(x)=`(x-cos36^@)(x-cos84^@)(x-cos156^@)` then coefficient of `x^2` is |
| Answer» Correct Answer - A | |
| 6. |
Calculate without using trigonometric tables:A. `4 cos 20^(@)-sqrt(3)cot20^(@)`B. `(2cos40^(@)-cos20^(@))/(sin20^(@))`C. `cos^(6)(pi)/(16)+cos^(6)(3pi)/(16)+cos^(6)(5pi)/(16)+cos^(6)(7pi)/(16)`D. `tan10^(@)-tan50^(@)+tan70^(@)` |
| Answer» Correct Answer - A::B::C::D | |
| 7. |
Prove that `tan 70^(@)=cot70^(@)++2cot40^(@)` |
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Answer» LHS `=tan 70^(@)=tan(20^(@)+50^(@))=(tan20^(@)+tan 50^(@))/(1-tan20^(@)tan.50^(@))` or `tan70^(@)-tan20^(@)tan50^(@)tan70^(@)=tan20^(@)+tan50^(@)` or `tan70^(@)= tan70^(@)tan50^(@)tan20^(@)+tan20^(@)+tan50^(@)=2tan 50^(@)+tan20^(@)` `=cot70^(@)+2cot40^(@)=RHS`. |
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