Question

please its urjunt please help me​

Answer 1

$\large{\pmb{\purple{\sf Solution :-}}}$

Your required answers are marked with green.

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

• Now take the LCM

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

$: \implies \dfrac{ \green2}{1 - { \sin}^{2} \theta }$

• We know that:-

$\leadsto \underline {\boxed { \pink{{ \frak{ 1 - { \sin}^{2} \theta = { \cos}^{2} \theta }}}}}$

$\: \: \: \: \: \dag \underline{ \frak{substituting \: this \: value \: in \: the \: equation}}$

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

• we know that ... 1/cosØ = secØ

$: \implies \boxed { \pink{{\frak{2 { \sec}^{2} \theta }}}} \bigstar$

I hope it's beneficial :D

Answer 2

To Prove :-

$\frac{1}{1 + sinθ} + \frac{1}{1 - sinθ} = 2 {sec}^{2}θ$

Used Concepts :-

• In two fractions either they are under subtraction or addition , if we don't know the denominator ' s values i. e it is in a variable or other so , the product of denominators is the L.C.M Eg :- If three fractions are in addition and their denominators are x , y and z . So , the L.C.M of denominators is x . y . z respectively.
• Sin² x + Cos² x = 1
• ( a + b ) ( a - b ) = a² - b² .

Solution :-

Let L.H.S i.e :-

$\frac{1}{ 1 + sinθ} + \frac{1}{1 - sinθ}$

Taking L.C.M we get ,

$\frac{(1 - sinθ) + (1 + sinθ)}{(1 + sinθ) \times (1 - sinθ)}$

Applying a² - b² formula to the Denominator ,

$\frac{1 - sinθ + sinθ + 1}{1 - {sin}^{2} θ}$

Sinθ will be cancelled from the numerator and as we know that 1 = Sin² x + Cos²x

So ,

$\frac{2}{ {cos}^{2}θ}$

$2 \times \frac{1}{ {cos}^{2}θ }$

As we know that Sec x = 1/Cos x . So ,

$2 {sec}^{2} θ$

Therefore , L.H.S = R.H.S

Hence , Proved !