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Principal Solution of Trigonometric Equation

Prove that ...

Question

Prove that $sin A (1 + tan A) + cos A (1 + cot A) = sec A + c o s e c A$

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Solution

$s i n A (1 + t a n A) + c o s A (1 + c o t A)$
$= s i n A + s i n A t a n A + c o s A + c o s A c o t A$
$= s i n A + s i n A \frac{s i n A}{c o s A} + c o s A + c o s A \frac{c o s A}{s i n A}$ $[∵ t a n A = \frac{s i n A}{c o s A}, c o t A = \frac{c o s A}{s i n A}]$
$= s i n A + \frac{s i n^{2} A}{c o s A} + c o s A + \frac{c o s^{2} A}{s i n A}$
$= \frac{s i n^{2} A c o s + s i n^{3} A + c o s^{2} A s i n A + c o s^{3} A}{s i n A c o s A}$
$= \frac{s i n A c o s A (s i n A + c o s A) s i n^{3} A + c o s^{3} A}{s i n A c o s A}$
$= \frac{s i n A c o s A (s i n A + c o s A) (s i n^{2} A + c o s^{2} A - c o s A s i n A)}{s i n A c o s A}$
$[∵ a^{3} + b^{3 - (a + b) (a^{2} - a b + b^{2})}]$
$= \frac{(s i n A + c o s A) s i n A c o s A + s i n^{2} A + c o s^{2} A - s i n a + c o s A}{s i n a c o s A}$
$= \frac{(s i n A + c o s A) .1}{s i n a c o s A} [∵ s i n^{^{2} a + c o s^{2} A = 1}]$
$= \frac{s i n A}{s i n A c o s A} + \frac{c o s A}{s i n A c o s A}$
$= \frac{1}{C o s A} + \frac{1}{s i n a}$
=secA+cosec A.

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Q. Prove that SinA(1+tanA)+cosA(1+cotA)=secA+coecA

sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A .

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

Q. The expression

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is equivalent to

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