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General Solution of Trigonometric Equation

Prove that ...

Question

Prove that $\frac{c o s A c o s e c A - s i n A s e c A}{c o s A + s i n A} = c o s e c A - s e c A$

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Solution

LHS $= \frac{c o s A c o s e c A - s i n A s e c A}{c o s A + s i n A}$

$= \frac{{cos}^{2} A - {sin}^{2} A}{sin A cos A (cos A + sin A)}$

$= \frac{(cos A - sin A) (cos A + sin A)}{sin A cos A (cos A + sin A)}$

$= \frac{cos A - sin A}{sin A cos A}$

$= \frac{1}{s i n A} - \frac{1}{c o s A}$

$= c o s e c A - sec A$

$= R H S$

Hence proved

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Similar questions

Q. Prove the following trigonometric identities.

(i)

\frac{\cos A cosec A - \sin A \sec A}{\cos A + \sin A} = cosec A - \sec A

\frac{\sin A}{\sec A + \tan A - 1} + \frac{\cos A}{cosec A + \cot A - 1} = 1

\frac{c o s A - s i n A + 1}{c o s A + s i n A - 1} = c o s e c A + c o t A

\frac{1}{s i n A + c o s A + 1} + \frac{1}{s i n A + c o s A - 1} = s e c A + c o s e c A

Q. Prove that :

(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A .

Q. Prove the following identity :

(s i n A + s e c A)^{2}

(c o s A + c o s e c A)^{2}

(1 + s e c A . c o s e c A)^{2}

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General Solution of Trigonometric Equation

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