Advertisements
Advertisements
Question
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Advertisements
Solution
LHS = `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2`
= `tan A/(sec^2 A)^2 + cot A/(cosec^2 A)^2`
= `sin A/cos A xx cos^2 A xx cos^2 A + cos A/sin A xx sin^2 A xx sin^2 A`
= sin A.cos3A + sin3A.cos A
= sin A cos A (cos2 A + sin2 A)
= sin A. cos A x 1
= sin A. cos A
= RHS
Hence proved.
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 identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
