( 1 + \, cosec\, )^2 ( - )^2 ( 1 + ) ( + )^2 + ( + cosec\, )^2, 0° < θ < 90°, is equal to:

Option A. 1 + sinθ. Answer & Solution Answer: Option A Solution: & ( 1 + \, cosec\, )^2 ( - )^2 ( 1 + ) ( + )^2 + ( + cosec\, )^2 & = ( 1 + 1 )^2 ( 1 - )^2 ( 1 + ) ( + 1 )^2 + ( + 1 )^2 & = ( + 1 )^2 ( 1 - )^2 ( 1 + ) ( + 1 )^2 + ( + 1 )^2 & = ( + 1 )^2 ( 1 - )^2 ( 1 + ) ( + 1 )^2 ( 1 ^2 + 1 ^2 ) &…

( 1 + \, cosec\, )^2 ( - )^2 ( 1 + ) ( + )^2 + ( + cosec\, )^2, 0° <…

Answer: Option A

Solution:

\frac{( 1 + sec θ cosec θ ) ^{2} ( sec θ - tan θ ) ^{2} ( 1 + sin θ )}{( sin θ + sec θ ) ^{2} + ( cos θ + cosec θ ) ^{2}} = \frac{( 1 + \frac{1}{c o s θ s i n θ} ) ^{2} ( \frac{1}{c o s θ} - \frac{s i n θ}{c o s θ} ) ^{2} ( 1 + sin θ )}{( sin θ + \frac{1}{c o s θ} ) ^{2} + ( cos θ + \frac{1}{s i n θ} ) ^{2}} = \frac{( \frac{c o s θ s i n θ + 1}{c o s θ s i n θ} ) ^{2} ( \frac{1 - s i n θ}{c o s θ} ) ^{2} ( 1 + sin θ )}{( \frac{c o s θ s i n θ + 1}{c o s θ} ) ^{2} + ( \frac{c o s θ s i n θ + 1}{s i n θ} ) ^{2}} = \frac{( \frac{c o s θ s i n θ + 1}{c o s θ s i n θ} ) ^{2} ( \frac{1 - s i n θ}{c o s θ} ) ^{2} ( 1 + sin θ )}{( cos θ sin θ + 1 ) ^{2} ( \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ} )} = \frac{( \frac{c o s θ s i n θ + 1}{c o s θ s i n θ} ) ^{2} ( \frac{1 - s i n θ}{c o s θ} ) ^{2} ( 1 + sin θ )}{( cos θ sin θ + 1 ) ^{2} ( \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ} )} = (\frac{1 - sin θ}{cos θ})^{2} (1 + sin θ) = \frac{( 1 - sin θ ) ^{2} ( 1 + sin θ )}{cos ^{2} θ} = \frac{( 1 - sin θ ) ( 1 + sin θ ) ( 1 + sin θ )}{( 1 - sin ^{2} θ )} = (1 + sin θ)