Question

prove:
1/1+sin thita + 1/1-sin thita = 2 sec square thita​

Answer 1

Step-by-step explanation:

LHS=(1-sin theta+1+sin theta)/(1-sin^2theta)

=2/cos^2theta

=2sec^2 theta=RHS

Answer 2

To prove :-

$\tt { : \implies\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} = 2 \sec ^{2} \theta}$

Solution :-

» Consider LHS

$\tt { : \implies\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} }$

» Taking LCM

$\tt { : \implies\dfrac{1 - \sin \theta + 1 + \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)} }$

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

» Apply formula (A+B)(A-B)=A²-B²

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

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

» Apply formula 1-sin²θ = cos²θ

$\tt { : \implies\dfrac{2}{\cos^{2} \theta } }$

» Apply formula 1/cos²θ = sec²θ

$\tt { : \implies \purple{ 2 \sec ^{2} \theta } }$

» LHS = RHS

Hence proved