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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

tan A/1-cotA+...

Question

$\frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = (s e c A c o s e c A + 1)$

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Solution

$L H S = \frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = \frac{(s i n A / c o s A)}{(1 - \frac{c o s A}{s i n A})} + \frac{(c o s A / s i n A)}{1 - \frac{s i n A}{c o s A}}$

$\frac{s i n A}{c o s A \frac{(s i n A - c o s A)}{s i n A}} + \frac{c o s A}{s i n A \frac{(c o s A - s i n A)}{c o s A}}$

$= \frac{s i n^{2} A}{c o s A (s i n A - c o s A)} + \frac{c o s^{2} A}{s i n A (c o s A - s i n A)} = \frac{s i n^{3} A - c o s^{3} A}{c o s A s i n A (s i n A - c o s A)}$

$= \frac{(s i n A - c o s A) (s i n^{2} A + c o s^{2} A + s i n A c o s A)}{c o s A s i n A (s i n A - c o s A)}$ [Using $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)]$

$= \frac{1 + s i n A c o s A}{s i n A c o s A} [∵ s i n^{2} A + c o s^{2} A = 1]$

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

=sec A cosec A+1

$[∵ \frac{1}{c o s A} = s e c A, \frac{1}{s i n A} = c o s e c A]$

=RHS

Hence proved.

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Similar questions

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = sec A c o s e c A + 1

tan A / 1- cot A + cot A / 1- tan A = 1+ sec A cosec A.

Prove that :

tanA/1-cotA + cotA/1-tanA = secAcosecA+1

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = 1 + sec A c o s e c A

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

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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

Standard XII Mathematics

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