Prove sin -1 - cos - - cos ec - - -

LHS = sinθ1 cosθ = sinθ( 1 + cosθ)1 cos^2θ = ( 1 + cosθ)sinθsin^2θ = 1 + cosθsinθ = cos ecθ #160; + θ #10; #10;= #10;RHSHe ...

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Sin(A+B)Sin(A-B)

Prove sin θ...

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Prove $\frac{sin θ}{1 - cos θ} = cos e c θ + cot θ$

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Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

\frac{cosec θ + cot θ}{cosec θ - cot θ} = \frac{81}{49}

\frac{cos θ + sin θ}{sin θ - cos θ}

If $(c o s e c θ - s i n θ) = a^{3}$ and $(s e c θ - c o s θ) = b^{3}$ , prove that

$a^{2} b^{2} (a^{2} + b^{2}) = 1$

\frac{cot θ - cos θ}{cot θ + cos θ} = \frac{cosec θ - 1}{cosec θ + 1}

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