1-1-sin- -1-1-sin - - 2sec2-

LHS=1/(1+sin θ) +1/1 sin θ = (1−sinθ)+(1+sinθ)/(1+sinθ)(1−sinθ) = 1−sinθ+1+sinθ/(1+sinθ)(1−sinθ) = 2/1−sin^2 θ = 2/cos^2 θ =2 ×1/cos^2 θ =2 sec^2 θ =RHS ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

1/1+sin θ +1/...

Question

$\frac{1}{(1 + s i n θ)} + \frac{1}{1 - s i n θ} = 2 s e c^{2} θ$

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Solution

LHS= $\frac{1}{(1 + s i n θ)} + \frac{1}{1 - s i n θ}$

$= \frac{(1 - s i n θ) + (1 + s i n θ)}{(1 + s i n θ) (1 - s i n θ)}$

$= \frac{1 - s i n θ + 1 + s i n θ}{(1 + s i n θ) (1 - s i n θ)}$

$= \frac{2}{1 - s i n^{2} θ}$

$= \frac{2}{c o s^{2} θ}$

$= 2 \times \frac{1}{c o s^{2} θ}$

$= 2 s e c^{2} θ$

=RHS

Suggest Corrections

1(1+sin θ)+11−sin θ=2sec2θ

LHS=1(1+sin θ)+11−sin θ =(1−sinθ)+(1+sinθ)(1+sinθ)(1−sinθ) =1−sinθ+1+sinθ(1+sinθ)(1−sinθ) =21−sin2θ =2cos2θ =2×1cos2θ =2 sec2θ =RHS

$\frac{1}{(1 + s i n θ)} + \frac{1}{1 - s i n θ} = 2 s e c^{2} θ$