Prove that 11 - cos- - 11 - cos- - 2cosec2-

Solving LHS 11+cosθ+11 cosθ=1 cosθ+1+cosθ(1+cosθ)(1 cosθ) (cross multiplying) 1 cosθ+1+cosθ(1+cosθ)(1 cosθ)=21 cos ^2θ 21 cos ^2θ=2sin^2θ=2cosec^2θ ...

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Standard XII

Mathematics

Circular Measurement of Angle

Prove that ...

Question

Prove that $\frac{1}{1 + cos θ} + \frac{1}{1 - cos θ} = 2 {cosec}^{2} θ$ .

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Solution

Solving LHS
$\frac{1}{1 + cos θ} + \frac{1}{1 - cos θ} = \frac{1 - cos θ + 1 + cos θ}{(1 + cos θ) (1 - cos θ)}$
(cross multiplying)

$\frac{1 - cos θ + 1 + cos θ}{(1 + cos θ) (1 - cos θ)} = \frac{2}{1 - {cos}^{2} θ}$

$\frac{2}{1 - {cos}^{2} θ} = \frac{2}{{sin}^{2} θ} = 2 {cosec}^{2} θ$
$L H S = R H S$
Hence Proved.

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Prove that 11+cosθ+11−cosθ=2cosec2θ.

Solving LHS11+cosθ+11−cosθ=1−cosθ+1+cosθ(1+cosθ)(1−cosθ)(cross multiplying)1−cosθ+1+cosθ(1+cosθ)(1−cosθ)=21−cos2θ21−cos2θ=2sin2θ=2cosec2θLHS=RHSHence Proved.

Prove that $\frac{1}{1 + cos θ} + \frac{1}{1 - cos θ} = 2 {cosec}^{2} θ$ .