If coto=7/8, then find value of (1-sino)(1+sino)/(1-coso)(1+coso)

1.	If coto=7/8, then find value of (1-sino)(1+sino)/(1-coso)(1+coso)
Answer» Properties used :-) $(x - y)(x + y) = {x}^{2} - {y}^{2} \\ \\ 1 - {sin}^{2} \theta = {cos}^{2} \theta \\ \\1 - {cos}^{2} \theta = {sin}^{2} \theta \\ \\ \frac{ {cos}^{2}\theta} { {sin}^{2}\theta} = {cot}^{2} \theta$ ln this question we have GIVEN that $cot \theta = \frac{7}{8}$ And we have to find the VALUE of $\Large{ \frac{(1 - sin\theta)(1 + sin \theta)}{(1 - cod\theta)(1 + cos\theta)} }$ So, $\large{ \frac{(1 - sin\theta)(1 + sin \theta)}{(1 - cos\theta)(1 + cos\theta)} } \\ \\ = \frac{ {1}^{2} - {sin}^{2} \theta}{ {1}^{2} - {cos}^{2} \theta} \\ \\ = \frac{1 - {sin }^{2} \theta }{1 - {cos}^{2} \theta} \\ \\ = \frac{ {cos}^{2} \theta}{ {sin}^{2} \theta} \\ \\ = {cot}^{2} \theta \\ \\$ $= {( \frac{7}{8} )}^{2} \\ \\ = \frac{49}{64}$ Which is the REQUIRED ANSWER.

Answer»

Properties used :-)

$(x - y)(x + y) = {x}^{2} - {y}^{2} \\ \\ 1 - {sin}^{2} \theta = {cos}^{2} \theta \\ \\1 - {cos}^{2} \theta = {sin}^{2} \theta \\ \\ \frac{ {cos}^{2}\theta} { {sin}^{2}\theta} = {cot}^{2} \theta$

ln this question we have GIVEN that

$cot \theta = \frac{7}{8}$

And we have to find the VALUE of

$\Large{ \frac{(1 - sin\theta)(1 + sin \theta)}{(1 - cod\theta)(1 + cos\theta)} }$

So,

$\large{ \frac{(1 - sin\theta)(1 + sin \theta)}{(1 - cos\theta)(1 + cos\theta)} } \\ \\ = \frac{ {1}^{2} - {sin}^{2} \theta}{ {1}^{2} - {cos}^{2} \theta} \\ \\ = \frac{1 - {sin }^{2} \theta }{1 - {cos}^{2} \theta} \\ \\ = \frac{ {cos}^{2} \theta}{ {sin}^{2} \theta} \\ \\ = {cot}^{2} \theta \\ \\$