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1.

सिद्ध कीजिए `(cos theta)/(1-tan theta)+(sin theta)/(1-cot theta)=cos theta+sin theta.`

Answer» यहाँ `L.H.S. =(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 theta)/(cos theta-sin theta))/(cos theta)+((sin theta)/(sin theta-cos theta))/(sin theta)=(cos ^(2)theta)/(cos theta-sintheta)+(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=R.H.S.`
Discussion
2.

सिद्ध कीजिए `(tan theta)/(1-cot theta)+(cot theta)/(1-tan theta)=1+tan theta+cot theta=1+sec thetacosec theta.`

Answer» यहाँ `L.H.S. =(tan theta)/(1-cot theta)+(cot theta)/(1-tan theta)`
`=(tan theta)/(1-(1)/(tantheta))+((1)/(tan theta))/(1-tan theta)=(tan theta)/((tan theta-1)/(tan theta))+(1)/(tan theta(1-tan theta))`
`=(tan ^(2)theta)/(tan theta^(-1))+(1)/(tan theta(1-tan theta))=(tan ^(2)theta)/(tan theta-1)-(1)/(tan theta(tan theta-1))`
`=(tan ^(3)theta-1)/(tan theta(tan theta-1))=((tan theta-1)(tan ^(2)theta+tan theta+1))/(tan theta(tan theta-1))`
`(tan ^(2)theta+tan theta+1)/(tan theta)=(tan ^(2)theta)/(tan theta)+(tan theta)/(tan theta)+(1)/(tan theta)`
`=tan theta+1+cot theta=(1+tan theta+cot theta)`
`=1+(sin theta)/(cos theta)+(cos theta)/(sin theta)=1+(sin ^(2)theta+cos ^(2)theta)/(sin thetacos theta)=1+(1)/(sin thetacos theta)`
`=1+cosec theta sec theta=R.H.S.`
Discussion
3.

सिद्ध कीजिए `(1+tan^(2)theta)/(1+cot^(2)theta)=((1-tan theta)/(1-cottheta))^(2)=tan ^(2)theta.`

Answer» यहाँ L.H.S. `=(1+tan ^(2)theta)/(1+cot ^(2)theta)=(sec^(2)theta)/(cosec ^(2)theta)`
`=((1)/(cos ^(2)theta))/((1)/(sin ^(2)theta))=(1)/(cos ^(2)theta)xx(sin^(2)theta)/(1)=tan ^(2)theta=R.H.S.`
अब मध्य पद `=((1-tan theta)/(1-cot theta))^(2)`
`=((1-(sin tan theta)/(cos theta))/(1-(cos theta)/(sin theta)))^(2)=(((cos theta-sin theta)/(cos theta))/((sin theta-cos theta)/(sin theta)))=((cos theta-sin theta)/(cos theta)xx(sin theta)/(sin theta-cos theta))^(2)`
`=((-sin theta)/(cos theta))^(2)=(-tantheta)^(2)=tan ^(2)theta=R.H.S.`
Discussion
4.

सिद्ध कीजिए `(1+sin theta)/(1-sin theta)=(sectheta+tan theta)^(2).`

Answer» यहाँ `L.H.S.=((1+sin theta))/((1-sintheta))=((1+sin theta))/((1-sintheta))xx((1+sin theta))/((1+sintheta))`
`=((1+sin theta)^(2))/((1-sin ^(2)theta))=((1+sin ^(2)theta+2sin theta))/(cos ^(2)theta)=((1)/(cos ^(2)theta)+(sin^(2)theta)/(cos ^(2)theta)+(2sintheta)/(cos ^(2)theta))`
`=(sec ^(2)theta+tan ^(2)theta+2 sectheta tan theta)=(sectheta+tan theta)^(2)=R.H.S.`
Discussion
5.

सिद्ध कीजिए `sin theta(1+tan theta)+cos theta(1+cot theta)=sectheta+cosec theta. `

Answer» यहाँ `L.H.S. =sin theta(1+tan theta)+cos (1+cos theta)`
`=sin theta(1+(sin theta)/(costheta))+cos theta(1+(cos theta)/(sin theta))`
`=sin theta((cos theta+sin theta)/(cos theta))+cos theta((sin theta+cos theta)/(sin theta))`
`=(sin^(2)(cos theta+sin theta)+cos ^(2)theta(sin theta+cos theta))/(cos thetasin theta)`
`=((sin ^(2)theta+cos ^(2))*(sin theta+cos theta))/(cos theta*sin theta)`
`=(sin theta)/(cos theta*sin theta)+(cos theta)/(cos theta*sin theta)=(1)/(cos theta)+(1)/(sin theta)`
`=sec theta+cosec theta=R.H.S.`
Discussion
6.

सिद्ध कीजिए `(sin ^(2)theta)/(cos ^(2)theta)+(cos ^(2)theta)/(sin ^(2)theta)=(1)/(sin ^(2)thetacos ^(2)theta)-2`

Answer» यहाँ `L.H.S. =(sin ^(2)theta)/(cos ^(2)theta)+(cos ^(2)theta)/(sin ^(2)theta)=(sin^(4)theta+cos ^(4)theta)/(sin^(2)thetacos ^(2)theta)`
`=((sin ^(2)theta)+(cos ^(2)theta)^(2)+2 sin^(2)thetacos ^(2)theta-2sin ^(2)thetacos ^(2)theta)/(sin ^(2)thetacos ^(2)theta)`
`=((sin^(2)theta+cos ^(2)theta)^(2)-2sin ^(2)thetacos ^(2)theta)/(sin ^(2)thetacos ^(2)theta)`
`=(1-2 sin ^(2)thetacos ^(2)theta)/(sin ^(2)thetacos ^(2)theta)=(1)/(sin ^(2)thetacos ^(2)theta)-2=R.H.S.`
Discussion
7.

यदि `cosec theta+cot theta=m,` सिद्ध कीजिए कि `(m^(2)-1)/(m^(2)+1)=cos theta.`

Answer» यहाँ `cosec theta +cot theta=m`
`implies(cosec theta+cot theta)^(2)=m^(2)`
`impliescosec ^(2)theta+cot ^(2)theta+2cosec theta cot theta=m^(2)`
यहाँ, `L.H.S.=(m^(2)-1)/(m^(2)+1)=(cosec ^(2)theta+cot^(2)theta+2 cosec theta cottheta-1)/(cosec ^(2)theta+cot^(2)theta+cosec theta cot theta+1)`
`=(cot ^(2)theta+cot ^(2)theta+2 cosec theta cot theta)/(cosec ^(2)+cosec ^(2)theta+2cosec theta cot theta)`
`=(2cot^(2)theta+2cosec thetacot theta)/(2 cosec ^(2)theta+2 cosec theta cot theta)=(2cot theta(cot theta+cosec theta))/(2 cosec theta(cosec theta+cot theta))=(cot theta)/(cosec theta)`
`=((cos theta)/(sin theta))/((1)/(sin theta))=(cos theta)/(sin theta)xx(sin theta)/(1)=cos theta=R.H.S.`
Discussion
8.

सिद्ध कीजिए कि `sqrt((1+cos theta )/(1-cos theta))=cosec theta+cot theta`

Answer» यहाँ `L. H. S. =sqrt((1+cos theta)/(1-cos theta))=sqrt((1+cos theta)/(1-cos theta)xx(1+cos theta)/(1+cos theta))`
`=sqrt(((1+cos theta)^(2))/(1-cos ^(2)theta))=sqrt(((1+cos theta)^(2))/(sin ^(2)theta))=sqrt(((1+cos theta)/(sin theta))^(2))=(1+cos theta)/(sin theta)`
`=(1)/(sin theta)+(cos theta)/(sin theta)=cosec theta+cot theta=R.H.S.`
Discussion
9.

सिद्ध कीजिए `(cos ^(2)theta)/(1-tan theta)+(sin ^(3)theta)/(sin theta-cos theta)=1+sin thetacos theta`

Answer» यहाँ L.H.S.` =(cos ^(2)theta)/(1-tan theta)+(sin ^(3)theta)/(sin theta-cos theta)`
`=(cos ^(3)theta)/(cos theta-sin theta)-(sin^(3)theta)/(cos theta-sin theta)=(cos ^(3)theta-sin ^(3)theta)/(cos theta-sin theta)`
`=((cos theta-sin theta)(cos ^(2)theta+sin ^(2)theta+cos thetasintheta))/((cos theta-sin theta))`
`=1+sin theta cos theta=`R.H.S.
Discussion
10.

`DeltaABC` में `angleB=90^(@)` तथा `AB=BC=5,` सेमी, तब `sin A` `tanC` के मान ज्ञात कीजिए.

Answer» यहाँ `AC=sqrt(AB^(2)+BC^(2))=sqrt(5^(2)+5^(2))=sqrt50 =5sqrt2`
अब `sin A=(BC)/(AC)=(5)/(5sqrt2)=(1)/(sqrt2)`
तथा `tan C=(AB)/(BC)=5/5=1`
Discussion
11.

`(4)/(cot^(2)30^(@))+(1)/(sin ^(2)60^(@))-cos ^(2)45^(@)` का मान ज्ञात कीजिए.

Answer» यहाँ `(4)/(cos ^(2)30^(@))+(1)/(sin ^(2)60^(@))-cos ^(2)45^(@)`
`=(4)/((sqrt3))+(1)/(((sqrt3)/(2))^(2))-((1)/(sqrt2))^(2)=4/3+(1)/((3)/(4))-1/2=4/3+4/3-1/2=(8+8-3)/(6)=13/6.`
Discussion
12.

यदि `sin theta++cos theta=sqrt2sin(90^(@)-theta), ` तब `cot theta` का मान ज्ञात कीजिए.

Answer» यहाँ, `sin theta+cos theta=sqrt2sin (90^(@)-theta)impliessintheta+cos theta=sqrt2cos theta`
`impliessin theta=sqrt2cos theta-cos theta=(sqrt2-1)cos theta`
`implies(sin theta)/(cos theta)=((sqrt2-1)cos theta)/(cos theta)`
`implies tan theta=(sqrt2-1)impliescot theta=(1)/(sqrt2-1)`
`impliescot theta=((sqrt2+1))/((sqrt2-1)(sqrt2+1))=(sqrt2+1)/(2-1)=sqrt2+1`
Discussion
13.

निम्न के मान ज्ञात कीजिए- `(sin 15^(@)cos 75^(@)+cos15^(@)sin 75^(@))/(tan 5^(@)tan 30^(@)tan 35^(@)tan55^(@)tan 85^(@)).`

Answer» यहाँ, `(sin 15^(@)cos 75^(@)+cos15^(@)sin 75^(@))/(tan 5^(@)tan 30^(@)tan 35^(@)tan 55^(@)tan 85^(@))`
`=(sin 15^(@)cos (90^(@)-15^(@))+cos 15^(@)sin (90^(@)-15^(@)))/(tan 5^(@)""(1)/(sqrt3)tan 35^(@)tan (90^(@)-35^(@))*tan (90^(@)-5^(@)))`
`=(sin 15^(@)*sin 15^(@)+cos 15^(@)cos 15^(@))/((1)/(cos 5^(@))*(1)/(sqrt3)*(1)/(cot 35^(@))*cot 35^(@).cot 5^(@))=(sin ^(2)15^(@)cos ^(2)15^(@))/((1)/(sqrt3))=1xx(sqrt3)/(1)=sqrt3.`
Discussion
14.

`2(sin 43^(@))/(cos 47^(@))-(cot 37^(@))/(tan 53^(@))-sqrt2cos 45^(@)` का मान ज्ञात कीजिए.

Answer» यहाँ `2 (sin 43^(@))/(cos 47^(@))-(cot 37^(@))/(tan 53^(@))-sqrt2cos 45^(@)`
`=2(sin (90^(@)-47^(@)))/(cos 47^(@))-(cot (90^(@)-53^(@)))/(tan 53^(@))-sqrt2*(1)/(sqrt2)`
`=2(cos 47^(@))/(cos 47^(@))-(tan 53^(@))/(tan 53^(@))-1=2-1-1=0`
Discussion
15.

`sec 55^(@)sin 35^(@)+cos 35^(@)cosec 55^(@)-tan^(2)60^(@)` का मान ज्ञात कीजिए.

Answer» यहाँ `sec 55^(@)sin 35^(@)+cos 35^(@)cosec 55^(@)-tan^(2)60^(@)`
`=sec 55^(@)sin (90^(@)-55^(@))+cos (90^(@)-55^(@))cosec 55^(@)-(sqrt3)^(2)`
`=(1)/(cos 55^(@))*cos 55^(@)+sin55^(@)*(1)/(sin 55^(@))-3=1+1-3=-1`
Discussion
16.

सिद्ध कीजिए कि `(1-sin60^(@))/(cos 60^(@))=(1-tan 30^(@))/(1+tan 30^(@))`

Answer» बायाँ पक्ष `(1-sin60^(@))/(cos60^(@))=(1-(1)/(sqrt2))/(1/2)=(2-sqrt3)/(1)=2-sqrt3`
अब , बायाँ पक्ष `=(1-sin60^(@))/(cos 60^(@))=(1-(1)/(sqrt3))/(1+(1)/(sqrt3))=(sqrt3-1)/(sqrt3+1)=((sqrt3-1)xx(sqrt3-1))/((sqrt3+1)xx(sqrt3-1))`

`=((sqrt3-1)^(2))/((sqrt3)^(2)-(1)^(2))=(3-2sqrt3+1)/(2)=(4-2sqrt3)/(2)=2-sqrt3`
`implies` बायाँ पक्ष = दायाँ पक्ष
Discussion
17.

`tan 35^(@)tan 40^(@)tan 45^(@)tan 50^(@)tan 55^(@)` का मान ज्ञात कीजिए.

Answer» यहाँ, `tan 35^(@)tan40^(@)tan 45^(@)tan 50^(@)tan 55^(@)`
`=tan 35^(@)xxtan 40^(@)xx1xxtan (90^(@)-40^(@))tan (90^(@)-35^(@))`
`=tan35^(@)*tan40^(@)cot 40^(@)cot 35^(@)`

`=tan 35^(@)*tan 40^(@)xx(1)/(tan 40^(@))xx(1)/(tan 35^(@))=1`
Discussion
18.

सिद्ध कीजिए कि(i) `((sin 35^(@))/(cos 55^(@)))+((cos 55^(@))/(sin 35^(@)))-2cos60^(@)=1` (ii) `((sin47^(@))/(cos 43^(@)))+((cos 43^(@))/(sin 47^(@)))=4cos ^(2)45^(@)` (iii) `(sin ^(2)63^(@)+sin ^(2)27^(@))/(sec ^(2)20^(@)-cot^(2)70^(@))+2sin 36^(@)sin 42^(@). sec 48^(@)sec 54^(@)=3`

Answer» (i) L.H. S. `=((sin 35^(@))/(cos 55^(@)))+((cos 55^(@))/(sin 35^(@)))-2cos 60^(@)`
`=((cos (90^(@)-55^(@)))/(cos 55^(@)))^(2)+((cos (90^(@)-35^(@)))/(sin 35^(@)))-2cos 60^(@)`
`=((cos 55^(@))/(cos 55^(@)))+((sin 35^(@))/(sin 35^(@)))^(2)-2cos 60^(@)`
`=1^(2)+1^(2)-2xx1/2=1+1-1=1=R.H.S.`
(ii) `L.H.S.=((sin 47^(@))/(cos 43^(@)))^(2)+((cos 43^(@))/(sin 47^(@)))^(2)`
`=(sin (90^(@)-43^(@))/(cos 43^(@)))^(2)+((cos (90^(@)-47^(@)))/(sin47^(@)))=((cos 43^(@))/(cos 43^(@)))+((sin47^(@))/(sin 47^(@)))^(2)=1^(2)+1^(2)=2`
R.H.S. `=4cos ^(2)45^(@)=4xx((1)/(sqrt2))^(2)=4xx1/2=2`
अतः R.H.S. = L.H.S
(ii) L. H. S. `=(sin ^(2)63^(@)+sin^(2)(90-63)^(@))/(sec^(2)20^(@)-cso ^(2)(90-20)^(@))+2sin (90-54)^(@)sin (90-48)^(@). sec 48^(@)*sec 54^(@)`
`=(sin ^(2)63^(@)+cos ^(2)63^(@))/(sec ^(2)20^(@)-tan ^(2)20^(@))+2cos 54^(@)*cos 48^(@)*(1)/(cos 48^(@))*(1)/(cos 54^(@))`
`=1/1+2=3=` R.H.S.
Discussion
19.

निम्न का मान ज्ञात कीजिए- `(2 sin ^(2)60^(@)+1+2sin ^(2)27^(@))/(3cos ^(2)17^(@)-2+cos ^(2)73^(@))`

Answer» यहाँ `(2sin ^(2)63^(@)+1+2sin ^(2)73^(@))/(3cos ^(2)17^(@)-2+3cos ^(2)73^(@))=(2[sin ^(2)63^(@)+sin ^(2)(90^(@)-63^(@))]+1)/(3[cos ^(2)17^(@)+cos ^(2)(90^(@)-17^(@))]-2)`
`=(2[sin^(2)63^(@)+cos^(2)63^(@)]+1)/(3[cos ^(2)17^(@)+sin ^(2)17^(@)]-2)=((2xx1)+1)/((3xx1)-2)=3`
Discussion
20.

सिद्ध कीजिए `(cosec theta)/((cosec theta-1))+(cosec theta)/((cosec theta+1))= 2sec^(2)theta.`

Answer» यहाँ `L.H.S.=(cosec theta)/((cosec theta-1))+(cosec theta)/((cosec theta+1))`
`=(cosec (cosec theta+1)+cosec theta(cosec theta-1))/((cosec ^(2)theta-1))`
`=(2cosec ^(2)theta)/((1+cot^(2)theta-1))=(2cosec^(2)theta)/(cot ^(2)theta)=2cosec ^(2)thetatan ^(2)theta`
`=(2xx(1)/(sin ^(2)theta)xx(sin ^(2)theta)/(cos ^(2)theta))=(2)/(cos ^(2)theta)=2 sec^(2)theta=R.H.S.`
Discussion
21.

सिद्ध कीजिए कि `(1)/(1-sin theta)+(1)/(1+sin theta)=2 sec ^(2)theta`

Answer» यहाँ `L.H.S.=(1)/(1-sin theta)+(1)/(1+sin theta)`
`=(1+sintheta+1-sin theta)/((1-sin theta)(1+sin theta))=(2)/(1-sin ^(2)theta)=(2)/(1-sin^(2)theta)=2 sec^(2)theta=R.H.S.`
Discussion
22.

`cos 42^(@)-sin48^(@)` का मान ज्ञात कीजिए.

Answer» यहाँ `cos 42^(@)-sin48^(@)=cos (90^(@)-48^(@))-sin48^(@)`
`=sin (48^(@))-sin48^(@)" "[because cos (90^(@)-theta)=sin theta]`
`=sin 48^(@)-sin 48^(@)=0`
Discussion
23.

यदि `sin theta-cos theta=1/2,` तब `(1)/(sin theta+cos theta)` का मान ज्ञात कीजिए.

Answer» यहाँ `sin theta -cos theta =1/2 implies(sin theta-cos theta)^(2)=((1)/(2))^(2)`
` implies2sin theta*cos theta=3/4" "...(1)`
अब, `(sin theta+cos theta)^(2)=sin^(2)theta+cos ^(2)theta+2 sin theta cos theta=1+3/4`
`implies(sintheta+cos theta)=sqrt((7)/(4))`
`implies(1)/(sin theta+cos theta)=(2)/(sqrt7)=(2sqrt7)/(7)*`
Discussion
24.

सिद्ध कीजिए कि `sqrt((1+cos theta)/(1-cos theta))+sqrt((1-cos theta)/(1+cos theta))=2cosec theta. `

Answer» यहाँ `L.H.S. =sqrt((1+cos theta)/(1-cos theta))+sqrt((1-cos theta)/(1+cos theta))=(1+cos theta+1-cos theta)/(sqrt((1-cos theta)(1+cos theta)))`
`=(2)/(sqrt(1-cos^(2)theta))=(2)/(sqrt(sin^(2)theta))=(2)/(sin theta)`
`=2 cosec theta=R.H.S.`
Discussion
25.

यदि `cosec theta=2, ` तब `(1)/(tan theta)+(sin theta)/(1+cos theta)` का मान ज्ञात कीजिए.

Answer» हम जानते है कि
पाइथागोरस प्रमेय से, `cosec theta(" कर्ण")/(" लम्ब")=2/1` दिया है
`("आधार")^(2)=("कर्ण")^(2)-(" लम्ब")^(2)=4-1=3`
`implies "आधार"=sqrt3`
`therefore sin theta=1/2, tan theta =(1)/(sqrt3)` व `cos theta=(sqrt3)/(2)`
अब `(1)/(tan theta)+(sin theta)/(1+cos theta)=(1)/((1)/(sqrt(3)))+(1//2)/(1+(sqrt3)/(2))=sqrt3+((1)/(2))/((2+sqrt3)/(2))`
`=(sqrt3)/(1)+(1)/((2+sqrt3)).((2-sqrt3))/((2-sqrt3))=(sqrt3)/(1)+(2-sqrt3)/(4-3)`
`=2-sqrt3+sqrt3=2`
Discussion
26.

समीकरण `(cos ^(2)theta)/(cot ^(2)theta-cos ^(2)theta)=3` को `theta` के लिए हल कीजिए.

Answer» यहाँ, `(cos ^(2)theta)/(cot^(2)theta-cos ^(2)theta)`
`impliescos ^(2)theta=3 cot ^(2)theta-3 cos ^(2)theta`
`implies4 cos ^(2)theta=3cot^(2)theta`
`implies4cos ^(2)theta=(3cos ^(2)theta)/(sin ^(2)theta)implies4 sin ^(2)theta=(3 cos ^(2)theta)/(cos ^(2)theta)`
`impliessin ^(2)theta=3/4impliessin theta=(sqrt3)/(2)impliestheta=60^(@).`
Discussion
27.

सिद्ध कीजिए कि `1+(tan ^(2)theta)/(1+sec theta)=sec theta. `

Answer» यहाँ `L.H.S. =1+(tan ^(2)theta)/(1+sec theta)=1+(sec ^(2)theta-1)/(1+sec theta)`
`=1+((sec theta+1)(sec theta-1))/((1+sec theta))=1+sec theta-1=sec theta=R.H.S.`
Discussion
28.

सिद्ध कीजिए कि `(1-cos theta)/(sin theta)+(sin theta)/(1-cos theta)=2cosec theta. `

Answer» यहाँ L.H.S. `=(1-cos theta)/(sin theta)+(sin theta)/(1-cos theta)=((1-cos theta)^(2)+sin ^(2)theta)/(sin theta(1-cos theta))`
`=(1+cos ^(2)theta-2cos theta +sin ^(2)theta)/(sin theta(1-cos theta))`
`=(1+cos ^(2)theta+sin ^(2)theta-2cos theta)/(sin theta(1-cos theta))=(1+1-2 cos theta)/(sin theta(1-cos theta))`
`=(2-2costheta)/(sin theta(1-cos theta))=(2(1-cos theta))/(sin theta(1-cos theta))=(2)/(sin theta)=2 cosec theta=R.H.S.`
Discussion
29.

सिद्ध कीजिए कि `(tan theta+sin theta)/(tan theta-sin theta)=(sec theta+1)/(sec theta-1)`

Answer» यहाँ `L.H.S. =(tan theta+sin theta)/(tan theta-sin theta)`
`((sin theta)/(cos theta)+sintheta)/((sin theta)/(cos theta)-sin theta)=(sin theta((1)/(costheta)+1))/(sin theta((1)/(cos theta)-1))=((1)/(cos theta)+1)/((1)/(cos theta)-1)=(sec theta+1)/(sec theta-1)=R.H.S.`
Discussion
30.

समीकरण `(cos theta)/(1-sin theta)+(cos theta)/(1+sin theta)=4` को हल कीजिए.

Answer» दिया है-
`(cos theta)/(1+sin theta)+(cos theta)/(1+sin theta)=4` lt brgt `implies(cos theta(1+sintheta)+cos theta(1-sin theta))/((1-sin theta)(1+sin theta))`
`implies(cos theta(1+sin theta+1-sin theta))/(1-sin ^(2)theta)=4implies (cos theta(2))/(cos ^(2)theta)=4`
`impliescos theta=2/4=1/2impliestheta=60^(@).`
Discussion
31.

सिद्ध कीजिए कि `(1-sin theta)/(1+sin theta)=(sec theta-tan theta)^(2).`

Answer» यहाँ `L.H.S.=(1-sin theta)/(1+sin theta)=(1-sin theta)/(1+sin theta)xx(1-sin theta)/(1-sin theta)`
`=((1-sin theta)^(2))/(1-sin^(2)theta)=((1-sin theta)^(2))/(cos ^(2)theta)=((1-sin theta)/(cos theta))^(2)=((1)/(cos theta)-(sin theta)/(cos theta))^(2)`
`=(sec theta-tan theta)^(2)=R.H.S.`
Discussion
32.

सिद्ध कीजिए `sqrt((sec theta-1)/(sec theta-1))+sqrt((sec theta+1)/(sec theta-1))=2 cosec theta. `

Answer» यहाँ, L.H.S.`=sqrt((sec theta-1)/(sec theta+1))+sqrt((sec theta+1)/(sectheta-1))=(sqrt(sec theta-1))/(sqrt(sec theta+1))+(sqrt(sec theta+1))/(sqrt(sec theta-1))`
`=((sec theta-1)+(sec theta+a))/(sqrt((sec theta+1)(sec theta-1)))=(2 sec theta)/(sqrt(sec ^(2)theta-1))=(2 sec theta)/(sqrt(tan^(2)theta))`
`=(2 sec theta)/(tan theta)=2 sec theta cot theta`
`=((2)/(cos theta)xx(cos theta)/(sin theta))=(2)/(sin theta)=2 cosec theta=R.H.S.`
Discussion
33.

सिद्ध कीजिए `(sin theta)/(cot theta+cosec theta)=2+(sin theta)/(cot theta-cosec theta).`

Answer» यहाँ `(sin theta)/(cot theta+cosec theta)=2+(sin theta)/(cot theta-cosec theta).`
`implies(sin theta)/(cot theta+cosec theta)-(sin theta)/(cot theta-cosec theta)=2`
अब `L.H.S. =(sin theta)/(cot theta+cosec theta)-(sin theta)/(cot theta-cosec theta)=(sin theta)/(cottheta+cosec theta)+(sin theta)/(cosec theta-cot theta)`
`=sin theta{(1)/(cosec theta+cot theta)+(1)/(cosec theta-cot theta)}`
`=sin theta{(cosectheta-cot theta+cosec theta+cot theta)/(cosec ^(2)-cot^(2)theta)}`
`=sin theta(2 cosectheta)=2=R.H.S.`
Discussion
34.

सिद्ध कीजिए की `sin^(2)theta+(1)/(1+tan ^(2)theta)=1`

Answer» यहाँ `L.H.S.=1+tan ^(2)theta+(1)/(1+tan^(2)theta)=sin ^(2)theta+(1)/(sec^(2)theta)`
`=sin ^(2)theta+cos^(2)theta=1=R.H.S.`
Discussion
35.

सिद्ध कीजिए `(cos theta-sin theta+1)/(cos theta+sin theta-1)=cosec theta+cot theta`

Answer» यहाँ L.H.S. `(cos theta-sin theta+1)/(cos theta+sin theta-1)=((1+cos theta-sin theta)(cos theta+sin theta+1))/((cos theta+sin theta-1)(cos theta+sin theta+1))`
`=((1+cos theta)^(2)-sin ^(2)theta)/((cos theta+sin theta)^(2)-(1)^(2))=(1+cos ^(2)theta+2 cos theta-sin^(2)theta)/(cos ^(2)theta+sin^(2)theta+2 cos theta sin theta-1)`
`=(cos ^(2)theta+cos ^(2)theta+2 cos theta)/(2 sin theta cos theta)=(2 cos ^(2)theta+2 cos theta)/(2 sin theta cos theta)`
`=(2 cos theta(cos theta+1))/(2 sin theta cos theta)=(cos theta+1)/(sin theta)`
`=(cos theta)/(sin theta)+(1)/(sin theta)=cot theta+cosec theta=R.H.S.`
Discussion
36.

सिद्ध कीजिए `sec theta(1-sintheta)(sec theta+tan theta)=1.`

Answer» यहाँ L.H.S. `=sec theta (1-sin theta)(sec theta+tan theta)`
`=(1)/(cos theta)(1-sin theta)((1)/(cos theta)+(sin theta)/(cos theta))`
`=((1-sin theta)(1+sintheta))/(cos ^(2)theta)=(1-sin^(2)theta)/(cos ^(2)theta)=(cos ^(2)theta)/(cos ^(2)theta)=1=R.H.S.`
Discussion
37.

सिद्ध कीजिए `sin ^(6)theta+cos ^(6)theta=1-3 sin ^(2)theta cos ^(2)theta`

Answer» यहाँ L.H.S. `=sin ^(6)theta+cos ^(6)theta=(sin ^(2)theta)^(3)+(cos ^(2)theta)^(3)`
`=(sin ^(2)theta+cos ^(2)theta)^(3)-3sin ^(2)theta*cos ^(2)theta*(sin ^(2)theta+cos ^(2)theta)`
`=(1)^(3)-3sin theta cos ^(2)theta*1`
`=1-3 sin ^(2)theta*cos ^(2)theta=R.H.S.`
Discussion
38.

वह न्यूनकोण `theta ` ज्ञात कीजिए जिसके लिए `(cos theta-sin theta)/(cos theta+sin theta)=(1-sqrt3)/(1+sqrt3)`

Answer» यहाँ
`(cos theta-sin theta)/(cos theta+sin theta)=(1-sqrt3)/(1+sqrt3)`
तब योगन्त्रनुपात विधि से
`((costheta-sintheta)+(cos theta+sin theta))/((cos theta-sin theta)-(cos theta+sin theta))=((1-sqrt3)+(1+sqrt3))/((1-sqrt3)-(1+sqrt3))`
`implies(2 cos theta)/(-2sin theta)=(2)/(-2sqrt3)implies-costheta=-(1)/(sqrt3)`
अर्थत `cot theta=(1)/(sqrt3)`
`impliestan theta=sqrt3=tan 60^(@)`
अतः `theta=60^(@)`
Discussion
39.

यदि `tan(theta_(1)+theta_(2))=(tan theta_(1)+tan theta_(2))/(1-tantheta_(1)tan theta_(2)), theta_(1)` व `theta_(1)` न्यूनकोण है तब `theta_(1)+theta_(2)` का मान ज्ञात कीजिए यदि `tan theta_(1)=1/2` व `tan theta_(2)=1/3`

Answer» दिया है `tan (theta_(1)+theta_(2))=(tan theta_(1)+tan theta_(2))/(1-tan theta_(1)tan theta_(2))`
`(1/2+1/3)/(1-1/2xx1/3)=((3+2)/(6))/(1-1/6)=(5/6)/(5/6)=1" "(because tan theta_(1)=1/2` व `tan theta_(2)=1/2)`
`impliestan(theta_(1)+theta_(2))=1impliestheta_(1)+theta_(2)=45^(@)`
Discussion
40.

यदि `tan ""(5theta)/(2)=sqrt3` तथा `theta` एक न्यूनकोण है तब `2 theta ` का मान ज्ञात कीजिए.

Answer» यहाँ `tan ""(5theta)/(2)=sqrt3=tan 60^(@)" "(because tan 60^(@)=sqrt3)`
`implies(5 theta)/(2)=60^(@)impliestheta=(60xx2)/(5)=24^(@)`
अतः `2 theta=2xx24=48^(@)`
Discussion