If tan2 θ = 2 tan2 φ + 1, then cos 2 θ + sin2 φ is equal to

Question

If tan2 θ = 2 tan2 φ + 1, then cos 2 θ + sin2 φ is equal to (A) 0 (B)  2 sin2 φ (C)  - cos 2 φ (D) none of these.

Tardigrade Admin · Accepted Answer

cos 2 θ + sin2 φ = (1- tan2 θ/1+ tan2θ) + sin2 φ = (1-(2 tan2 φ + 1 )/1+(2 tan2 φ + 1 )) + sin2 φ = (- tan2φ/1+ tan2 φ) + sin2 φ = - ( tan2 φ/ sec2 φ) + sin2 φ = 0

Q. If $tan^{2} θ = 2 tan^{2} ϕ + 1,$ then $cos 2 θ + sin^{2} ϕ$ is equal to

0

$2 sin^{2} ϕ$

$- cos 2 ϕ$

none of these.

$cos 2 θ + sin^{2} ϕ = \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ} + sin^{2} ϕ$
$= \frac{1 - ( 2 t a n ^{2} ϕ + 1 )}{1 + ( 2 t a n ^{2} ϕ + 1 )} + sin^{2} ϕ$
$= \frac{- t a n ^{2} ϕ}{1 + t a n ^{2} ϕ} + sin^{2} ϕ = - \frac{t a n ^{2} ϕ}{s e c ^{2} ϕ} + sin^{2} ϕ = 0$

Q. If tan2θ=2tan2ϕ+1, then cos2θ+sin2ϕ is equal to

0

2sin2ϕ

−cos2ϕ

none of these.

cos2θ+sin2ϕ=1+tan2θ1−tan2θ​+sin2ϕ =1+(2tan2ϕ+1)1−(2tan2ϕ+1)​+sin2ϕ =1+tan2ϕ−tan2ϕ​+sin2ϕ=−sec2ϕtan2ϕ​+sin2ϕ=0

Q. If $tan^{2} θ = 2 tan^{2} ϕ + 1,$ then $cos 2 θ + sin^{2} ϕ$ is equal to

$2 sin^{2} ϕ$

$- cos 2 ϕ$

$cos 2 θ + sin^{2} ϕ = \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ} + sin^{2} ϕ$
$= \frac{1 - ( 2 t a n ^{2} ϕ + 1 )}{1 + ( 2 t a n ^{2} ϕ + 1 )} + sin^{2} ϕ$
$= \frac{- t a n ^{2} ϕ}{1 + t a n ^{2} ϕ} + sin^{2} ϕ = - \frac{t a n ^{2} ϕ}{s e c ^{2} ϕ} + sin^{2} ϕ = 0$