Solve this
Question:

$\frac{(\sec \theta-\tan \theta)}{(\sec \theta+\tan \theta)}=\left(1+2 \tan ^{2} \theta-2 \sec \theta \tan \theta\right)$

 

Solution:

$\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}$

$=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$

$=\frac{(\sec \theta-\tan \theta)^{2}}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$

$=\frac{\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta} \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right.$ and $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

$=\frac{1+\tan ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}{1} \quad\left(1+\tan ^{2} \theta=\sec ^{2} \theta\right)$

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

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.