Question

rs agg ex 13 A TRIGONOMETRIC IDENTITIES​

Answer 1

Question :-

→ prove :

$\frac{1}{ 1 + \sin \theta} + \frac{1}{1 - \sin \theta} = 2 { \sec}^{2} \theta.$

Step-by-step explanation :-

Solving LHS ,

$\sf = \frac{1}{ 1 + \sin \theta} + \frac{1}{1 - \sin \theta} . \\ \\ \sf = \frac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 - \sin \theta) (1 + \sin \theta)} . \\ \\ \sf = \frac{1 - \cancel{ \sin \theta}+ 1 + \cancel{\sin \theta}}{1 - { \sin}^{2} \theta } . \\ \\ \sf= \frac{2}{ { \cos}^{2} \theta} . \\ \\ \huge \pink{\boxed{ \it = 2 { \sec}^{2} \theta.}}$

LHS = RHS .

Hence, it is proved .

Answer 2

Answer

2sec²θ

STEPS

1/+sinθ1 + 1/−sinθ1 =2sec 2θ

Time to prove LHS

$\begin{lgathered}\sf = \frac{1}{ 1 + \sin \theta} + \frac{1}{1 - \sin \theta} \\ \\ \sf = \frac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 - \sin \theta) (1 + \sin \theta)}\\ \\ \sf = \frac{1 - \cancel{ \sin \theta}+ 1 + \cancel{\sin \theta}}{1 - { \sin}^{2} \theta } \\ \\ \sf= \frac{2}{ { \cos}^{2} \theta} \\ \\ \huge \red{\boxed{ \it = 2 { \sec}^{2} \theta}}\end{lgathered}$

$\huge \red{\boxed{ \Hence Proved( LHS=RHS) }}\end{lgathered}$