Prove the following trigonometric identities- 1-sec-sec-sin2-1-cos-

We know that sin^2θ + cos^2θ= 1 secθ = 1/cosθ LHS = 1+ secθ/secθ = 1+ 1/cosθ/1/cosθ = cosθ +1/cosθ× cosθ =cosθ +1 RHS = sin^2θ/1 cosθ = 1 cos^2θ/1 cosθ = ...

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Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

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

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Solution

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

$L H S = \frac{1 + s e c θ}{s e c θ}$

$= \frac{1 + \frac{1}{c o s θ}}{\frac{1}{c o s θ}}$

$= \frac{c o s θ + 1}{c o s θ} \times c o s θ$

$= c o s θ + 1$

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

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

$= \frac{(1 + c o s θ) (1 - c o s θ)}{1 - c o s θ}$

$= c o s θ + 1$

$L H S = R H S$

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Prove the following trigonometric identities: 1+sec θsec θ=sin2 θ1−cos θ

We know that sin2θ+cos2θ=1 secθ=1cosθ LHS=1+secθsecθ =1+1cosθ1cosθ =cosθ+1cosθ×cosθ =cosθ+1 RHS=sin2θ1−cosθ =1−cos2θ1−cosθ =(1+cosθ)(1−cosθ)1−cosθ =cosθ+1 LHS=RHS

Prove the following trigonometric identities:

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