Question

If tan^2 theta = 1 + tan^2alpha, Prove that sin^2 theta =1/2( 1 + sin^2alpha)

Answer 1

Correct question :

If $\tt {tan^2 \theta = 1 + 2tan^2 \alpha}$ , prove that $\tt{sin^2 \theta = \dfrac{1}{2}(1+ sin^2 \alpha)}$

Solution :

$\tt {tan}^{2} \theta = 1 + 2 {tan}^{2} \alpha$

Using identity ,

$\tt {tan}^{2} \theta + 1 = {sec}^{2} \theta \\ \\ \boxed{ \tt \therefore {tan}^{2} \theta = {sec}^{2}\theta - 1}$

$\tt {sec}^{2} \theta - 1 = 1 + 2( {sec}^{2} \alpha - 1) \\ \\ \implies \tt {sec}^{2} \theta - 1 = 1 + 2 {sec}^{2} \alpha - 2 \\ \\ \implies \tt {sec}^{2} \theta - 1 = 2 {sec}^{2} \alpha - 1 \\ \\ \tt \implies {sec}^{2} \theta = 2 {sec}^{2} \alpha$

Now , we know that

$\tt sec \theta = \dfrac{1}{cos \theta} \\ \\ \boxed{\implies \tt {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta } }$

$\tt {sec}^{2} \theta = 2 {sec}^{2} \alpha \\ \\ \implies \tt \frac{1}{ {cos}^{2} \theta } = 2 \left( \frac{1}{ {cos}^{2} \alpha } \right) \\ \\ \implies \tt \frac{1}{ {cos}^{2} \theta} = \frac{2}{ {cos}^{2} \alpha } \\ \\ \implies \tt {cos}^{2}\alpha = 2 {cos}^{2} \theta$

Using the identity :

$\tt {cos}^{2} \theta + {sin}^{2} \theta = 1 \\ \\ \boxed{ \therefore\tt {cos}^{2} \theta = 1 - {sin}^{2} \theta }$

$\tt{cos}^{2} \alpha = 2 {cos}^{2} \theta \\ \\ \implies \tt1 - {sin}^{2} \alpha = 2(1 - {sin}^{2} \theta) \\ \\ \implies \tt 1 - {sin}^{2} \alpha = 2 - 2 {sin}^{2} \theta \\ \\ \implies \tt 2 {sin}^{2} \theta = 2 - 1 + {sin}^{2} \alpha \\ \\ \implies \tt 2 {sin}^{2} \theta = 1 + {sin}^{2} \alpha \\ \\ \boxed{ \implies \tt {sin}^{2} \theta = \frac{1}{2} (1 + {sin}^{2} \alpha )}$

Hence proved