Advertisements
Advertisements
Question
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Advertisements
Solution
LHS = `1/((cosA + sinA) - 1) + 1/((cosA + sinA) + 1)`
= `(cosA + sinA + 1 + cosA + sinA - 1)/((cosA + sinA)^2 -1)`
= `(2(cosA + sinA))/(cos^2A + sin^2A + 2cosAsinA - 1)`
= `(2(cosA + sinA))/(1 + 2cosAsinA - 1) = (cosA + sinA)/(cosAsinA)`
= `cosA/(cosAsinA) + sinA/(cosAsinA)`
= `1/sinA + 1/cosA`
= cosecA + secA
APPEARS IN
RELATED QUESTIONS
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove that sin4A – cos4A = 1 – 2cos2A
