Prove the following trigonometric identities- cosec-sin-cosec-sin-cot2-cos2-

Taking LHS we have ( cosecθ + sinθ)( cosecθ sinθ) → ( cosec^2θ sin^2θ) → [1+ cot^2θ ( 1 cos^2θ)] → [1+ cot^2θ 1 + cos^2θ] → cot^2θ + cos^2θ = RHS He ...

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Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

$(c o s e c θ + s i n θ) (c o s e c θ - s i n θ) = c o t^{2} θ + c o s^{2} θ$

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Solution

Taking LHS we have

$(c o s e c θ + s i n θ) (c o s e c θ - s i n θ$ )

$\to$ $(c o s e c^{2} θ - s i n^{2} θ)$

$\to$ $[1 + c o t^{2} θ - (1 - c o s^{2} θ$ )]
$\to$ $[1 + c o t^{2} θ - 1 + c o s^{2} θ$ ]
$\to$ $c o t^{2} θ + c o s^{2} θ$
= RHS
Hence proved .

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Similar questions

Prove the following trignometric identities:

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

Q. Prove that

(c o sec θ + sin θ) (c o sec θ - sin θ) = {cot}^{2} θ + {cos}^{2} θ

Q. The solution of the trigonometric equation

\frac{{cos}^{2} θ}{{cot}^{2} θ - {cos}^{2} θ} = 3, 0^{\circ} < θ < 90^{\circ}

Q. Prove the following trigonometric identities.

cosec θ \sqrt{1 - \cos^{2} θ} = 1

Q. The trigonometric equation

\frac{1 - sin θ}{1 + sin θ}

is equal to
[Given,

{sin}^{2} θ + {cos}^{2} θ = 1

]

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Prove the following trigonometric identities: (cosec θ+sin θ)(cosec θ−sin θ)=cot2θ+cos2θ

Taking LHS we have (cosecθ+sinθ)(cosecθ−sinθ) → (cosec2θ−sin2θ) → [1+cot2θ−(1−cos2θ)] → [1+cot2θ−1+cos2θ] → cot2θ+cos2θ = RHS Hence proved .

Prove the following trigonometric identities:

$(c o s e c θ + s i n θ) (c o s e c θ - s i n θ) = c o t^{2} θ + c o s^{2} θ$

Taking LHS we have

$(c o s e c θ + s i n θ) (c o s e c θ - s i n θ$ )

$\to$ $(c o s e c^{2} θ - s i n^{2} θ)$

$\to$ $[1 + c o t^{2} θ - (1 - c o s^{2} θ$ )]
$\to$ $[1 + c o t^{2} θ - 1 + c o s^{2} θ$ ]
$\to$ $c o t^{2} θ + c o s^{2} θ$
= RHS
Hence proved .