Question

tan square theta by sec theta minus 1 whole square is equals to 1 + cos theta by 1 - cos theta​

Answer 1

Solution:

Given that ;

$\dfrac{ \tan^{2} \theta}{( \sec \theta - 1)^{2}} = \dfrac{(1 + \cos \theta}{(1 - \cos \theta)}$

Consider LHS,

$\dfrac{ \tan^{2} \theta }{( \sec \theta - 1) {}^{2} }$

We know that,

tan²θ = sec²θ - 1

$\dfrac{ \sec^{2}\theta - 1 }{( \sec \theta - 1) {}^{2} }$

Also, a² - b² = (a + b)(a - b)

$\frac{ \cancel{ (\sec \theta - 1)}( \sec \theta + 1)}{ \cancel{(\sec \theta - 1)}( \sec \theta - 1)} \\ \\ \frac{ \sec \theta + 1}{ \sec \theta - 1}$

We know that,

• secθ = 1/cosθ

$\implies \frac{ \left( \frac{1}{ \cos \theta} + 1 \right)}{\left( \frac{1}{ \cos \theta} - 1 \right)} \\ \\ \implies \frac{\left( \frac{1 + \cos \theta}{ \cos \theta} \right)}{\left( \frac{1 - \cos \theta}{ \cos \theta} \right)} \\ \\ \bf Cancelling \: the \: denominator : \\ \\ \implies \bf \frac{1 + cos \theta}{1 - cos \theta} = RHS$

Hence, it is proved!

Answer 2

Answer:

Mark me as brainliest

.... Hope u get ur answer...