Question

Prove question no. 42

Answer 1

On taking L.H.S. :

$\frac{ \cos \alpha }{1 - \tan \alpha } + \frac{ \sin\alpha }{1 - \cot \alpha } \\ \\ = > \frac{ \cos\alpha }{1 - \tan \alpha } + \frac{ \sin \alpha }{1 - \frac{1}{ \tan\alpha } } \\ \\ = > \frac{ \cos \alpha }{1 - \tan \alpha } + \frac{ \sin \alpha }{ \tan \alpha - 1 } \times \tan\alpha \\ \\ = > \frac{ \cos\alpha }{1 - \tan\alpha } - \frac{ \sin \alpha \tan \alpha }{1 - \tan\alpha } \\ \\ = > \frac{ \cos \alpha - \sin \alpha \tan\alpha }{1 - \tan \alpha } \\ \\ = > \frac{ \cos \alpha - \sin \alpha \times \frac{ \sin\alpha }{ \cos\alpha } }{1 - \frac{ \sin\alpha }{ \cos \alpha } } \\ \\ = > \frac{ { \cos}^{2} \alpha - { \sin }^{2} \alpha }{ \cos\alpha } \times \frac{ \cos \alpha }{ \cos\alpha - \sin\alpha } \\ \\ = > \cos \alpha + \sin \alpha \\ \\ = > \sin\alpha + \cos \alpha = </strong><strong>R.H.S</strong><strong>.</strong><strong>$

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Answer 2

Question: To Prove that;

$\boxed{\sf \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA} = sinA + cosA}$

Identities and Ratios used to solve the question.

$\Longrightarrow \ \sf tanA = \frac{sinA}{cosA}$

$\Longrightarrow \ \sf cotA = \frac{cosA}{sinA}$

$\Longrightarrow \ \sf (a-b) = -(b-a)$

$\Longrightarrow \ \sf a^2 - b^2 = (a+b)(a-b)$

Proof:

$\sf LHS = \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA}$

$\Longrightarrow \ \sf \dfrac{cosA}{1 - \frac{sinA}{cosA}} + \dfrac{sinA}{1 -  \frac{cosA}{sinA}}$

$\Longrightarrow \ \sf \dfrac{cosA}{\frac{cosA - sinA}{cosA}} + \dfrac{sinA}{\frac{sinA - cosA}{sinA}}$

Denominator is transferred to the numerator.

$\Longrightarrow \ \sf \dfrac{cos^2A}{cosA - sinA} + \dfrac{sin^2A}{sinA - cosA}$

Using the formula (a - b) = - (b + a) we get,

$\Longrightarrow \ \sf \dfrac{cos^2A}{cosA - sinA} - \dfrac{sin^2A}{cosA - sinA}$

$\Longrightarrow \ \sf \dfrac{cos^2A - sin^2A}{cosA - sinA}$

Using a² - b² = (a + b)(a - b) we get,

$\Longrightarrow \ \sf \dfrac{(cosA + sinA)(cosA - sinA)}{cosA - sinA}$

Cancelling like terms we get,

$\Longrightarrow \ \sf cosA + sinA$

LHS = RHS

Hence Proved.