Question

prove that (tan theta +sec theta -1 ).(tan theta +1+sec theta) = 2sintheta/1-sintheta

Answer 1

Question

$\rm \: : \implies( \tan\theta + \sec\theta - 1) \times ( \tan\theta + 1 + \sec\theta) = \dfrac{2 \sin\theta}{1 - \sin\theta }$

Solution:-

Given

$\rm \: : \implies( \tan\theta + \sec\theta - 1) \times ( \tan\theta + 1 + \sec\theta)$

We have to prove

$: \implies \dfrac{2 \sin\theta}{1 - \sin\theta }$

Use this trigonometry identity

$\to \tan\theta = \dfrac{ \sin \theta }{ \cos \theta}$

$\to \sec \theta= \dfrac{1}{ \cos \theta }$

Now put the value

$\rm : \implies \bigg( \rm \: \dfrac{ \sin\theta}{ \cos \theta } + \dfrac{1}{ \cos\theta } - 1 \bigg) \times \bigg(\rm \: \dfrac{ \sin\theta}{ \cos \theta } + \dfrac{1}{ \cos\theta } + 1 \bigg)$

Taking lcm

$\rm : \implies \bigg( \dfrac{ \sin\theta+ 1 - \cos \theta }{ \cos\theta } \bigg) \bigg( \dfrac{ \sin\theta+ 1 + \cos \theta }{ \cos\theta } \bigg)$

$\rm : \implies \dfrac{ ( \sin \theta - \cos \theta + 1)( \sin \theta + 1 + \cos \theta) }{ \cos {}^{2} \theta}$

$\rm \: : \implies \dfrac{ \sin {}^{2} \theta + \sin\theta + \cos \theta \times \sin\theta - cos \theta \times \sin\theta - \cos \theta - \cos {}^{2} \theta + \sin \theta + 1 + \cos \theta }{ \cos {}^{2} \theta }$

$\rm : \implies \dfrac{ \sin {}^{2} \theta - \cos {}^{2} \theta + \sin\theta - \cos \theta + \sin \theta + 1 + \cos \theta }{ \cos {}^{2} \theta}$

$\rm : \implies \dfrac{ \sin {}^{2} \theta- \cos {}^{2} \theta+ 2 \sin\theta+ 1 }{ \cos {}^{2} \theta }$

$\rm : \implies \dfrac{ \sin {}^{2} \theta + 1- \cos {}^{2} \theta+ 2 \sin\theta }{ \cos {}^{2} \theta }$

$\rm : \implies \dfrac{ \sin {}^{2} \theta + \sin {}^{2} \theta + 2 \sin\theta }{ 1 - \sin {}^{2} \theta }$

Use this identity

$\rm \to(a {}^{2} - b {}^{2} ) = ( a - b)(a + b)$

now

$\rm : \implies \dfrac{2 \sin {}^{2} \theta + 2 \sin\theta }{ (1 - \sin {}^{} \theta)(1 + \sin\theta) }$

$\rm : \implies \dfrac{2 \sin {}^{} \theta (1 + \sin\theta )}{ (1 - \sin {}^{} \theta)(1 + \sin\theta) }$

$\rm : \implies \dfrac{2 \sin {}^{} \theta \cancel{ (1 + \sin\theta )}}{ (1 - \sin {}^{} \theta) \cancel{(1 + \sin\theta) } }$

$\rm : \implies \dfrac{2 \sin {}^{} \theta }{ 1 - \sin {}^{} \theta }$

Hence proved