Prove the following trigonometric identities- 1-sin-2-1-sin-2-2-cos2-1-sin2-1-sin2-

We know that cos^2θ+ sin^2θ = 1 LHS = (1+sinθ)^2+(1 sinθ)^2/2cos^2θ = 1+2sinθ + sin^2 θ+1 2sinθ + sin^2 θ/2cos^2θ = 2+2sin^2θ/2cos^2θ = 2(1+sin^2θ) /2cos^2θ = 1 ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

Prove the fol...

Question

Prove the following trigonometric identities:

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

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Solution

We know that $c o s^{2} θ + s i n^{2} θ = 1$

$L H S = \frac{(1 + s i n θ)^{2} + (1 - s i n θ)^{2}}{2 c o s^{2} θ}$

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

$= \frac{2 + 2 s i n^{2} θ}{2 c o s^{2} θ}$

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

$= \frac{1 + s i n^{2} θ}{c o s^{2} θ}$

$= \frac{1 + s i n^{2} θ}{1 - s i n^{2} θ} = R H S$

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Prove the following trigonometric identities: (1+sin θ)2+(1−sin θ)22 cos2θ=1+sin2θ1−sin2θ

We know that cos2θ+sin2θ=1 LHS=(1+sinθ)2+(1−sinθ)22cos2θ =1+2sinθ+sin2θ+1−2sinθ+sin2θ2cos2θ =2+2sin2θ2cos2θ =2(1+sin2θ)2cos2θ =1+sin2θcos2θ =1+sin2θ1−sin2θ=RHS

Prove the following trigonometric identities:

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