Advertisements
Advertisements
प्रश्न
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Advertisements
उत्तर
L.H.S. = (1 – tan A)2 + (1 + tan A)2
= (1 + tan2A – 2 tan A) + (1 + tan2A + 2 tan A)
= 2(1 + tan2A)
= 2 sec2A
= R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that `(sec A)/(tan A + cot A) = sin A`.
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
