Question

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Answer 1

Question :

cosec α+ cotα = a

so prove that,cosα = $\dfrac{a {}^{2} - 1}{a {}^{2} + 1}$

${\purple{\boxed{\large{\bold{Formula's}}}}}$

•Trignometric Formula's:

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1

•Componendo and dividendo:

For any equal ratio

$\dfrac{a}{b} = \dfrac{c}{d}$

we can apply Componendo and dividendo, which gives

$\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

Solution :

Given : cosec α+ cotα = a

$\implies \frac{1}{ \sin \alpha } + \frac{ \cos \alpha }{ \sin \alpha } = a$

$\implies \dfrac{1 + \cos \alpha }{ \sin \alpha } = a$ $\implies1 + \cos \alpha = a \times \sin\alpha$

Now squaring on both sides

$\implies(1 + \cos \alpha ) {}^{2} = a {}^{2} (\sin {}^{2} \alpha )$

$\implies(1 + \cos\alpha ) {}^{2} = a {}^{2} (1 - \cos {}^{2} \alpha )$

$\implies(1 + \cos \alpha) {}^{2} = a {}^{2} (1 + \cos \alpha)(1 - \cos \alpha)$

$\implies \dfrac{(1 + \cos\alpha )}{(1 - \cos \alpha) } = a {}^{2}$

Now use Componendo and dividendo

$\implies \dfrac{(1 + \cos \alpha) + (1 - \cos \alpha) }{(1 + \cos \alpha ) - (1 - \cos \alpha ) } = \dfrac{a {}^{2} + 1 }{a {}^{2} - 1}$

$\implies \dfrac{2}{2 \cos \alpha } = \dfrac{a {}^{2} + 1}{a {}^{2} - 1}$

$\implies \dfrac{1}{ \cos\alpha } = \dfrac{a {}^{2} + 1}{a {}^{2} - 1 }$

$\implies \sec \alpha = \dfrac{a {}^{2} - 1 }{a {}^{2} + 1 }$

We know that secx =$\dfrac{1}{cosx}$

$\implies \cos \alpha = \dfrac{a {}^{2} - 1 }{a {}^{2} + 1}$

$\huge{\bold{ Hence\: Proved}}$