Advertisements
Advertisements
Question
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Solution
=`(1+ cot^2 theta ) sin ^2 theta`
=` cosec ^2 theta xx 1/ ( cosec^2 theta)`
=1
APPEARS IN
RELATED QUESTIONS
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`(1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)^2`
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
If x = a tan θ and y = b sec θ then
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
(1 – cos2 A) is equal to ______.
