Advertisements
Advertisements
Question
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Advertisements
Solution
LHS = `sin^2A(1 - sin^2B) - (1 - sin^2A)sin^2B`
= `sin^2A - sin^2A.sin^2B - sin^2B + sin^2A.sin^2B`
= `sin^2A - sin^2B` = RHS
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 the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`

From the figure find the value of sinθ.
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Evaluate:
`(tan 65°)/(cot 25°)`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
