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Principal Solution of Trigonometric Equation

Prove tan A...

Question

Prove $tan A - cot A = \frac{1 - 2 {cos}^{2} A}{sin A cos A}$

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Solution

Simplifying the LHS of $tan A - cot A = \frac{1 - 2 {cos}^{2} A}{cos A sin A}$ .

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

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

$= \frac{1 - {cos}^{2} A - {cos}^{2} A}{sin A cos A}$

$= \frac{1 - 2 {cos}^{2} A}{sin A cos A}$

$= R H S$

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Similar questions

Q. Prove that :

cot A - tan A = \frac{2 {cos}^{2} A - 1}{sin A cos A}

\frac{tan A}{(1 + {tan}^{2} A)^{2}} + \frac{cot A}{(1 + {cot}^{2} A)^{2}} = sin A cos A

\frac{tan A}{(1 + {tan}^{2} A)^{2}} + \frac{cot A}{(1 + {cot}^{2} A)^{2}} = sin A cos A

\frac{t a n A}{(1 + t a n^{2} A)^{2}} + \frac{c o t A}{(1 + c o t^{2} A)^{2}} = s i n A c o s A

tan A = \sqrt{2} - 1

, then prove that

sin A cos A = \frac{1}{2 \sqrt{2}}

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Principal Solution of Trigonometric Equation

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