Question

Plzz prove the above ques using trigonometric identities.....​

Answer 1

Question :-

$\frac{ \cos \alpha }{1 - \sin \alpha } = \frac{1 + \sin\alpha }{ \cos \alpha }$

Solution :-

$\frac{ \cos\alpha }{1 - \sin \alpha } = \frac{1 + \sin\alpha }{ \cos\alpha }$

Now do cross Multiplication

$\cos {}^{2} \alpha = 1 - \sin {}^{2} \alpha$

$∵ \: \cos {}^{2} \alpha + \sin {}^{2} \alpha = 1 \\ \: \: \: \: \: \: cos {}^{2} \alpha \: \: \: = 1 - \sin {}^{2} \alpha \\ \\ \\ ∴ \cos {}^{2} \alpha = \cos {}^{2} \alpha \\ \\ \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: lhs = rhs \: \: \: \: \: \: (proved)$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

$\sf\dfrac{cos\theta}{1 - sin\theta} = {1+ sin \theta}{cos\theta}$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Answer}}}}}}}}$

$\sf\dfrac{cos\theta}{1 - sin\theta} = \dfrac{1+ sin \theta}{cos\theta}$

• cross multiplication

$\sf cos \theta \times cos \theta = (1-sin\theta)(1+cos\theta)$

${\bold{\blue{\boxed{\bf{ (a+b)(a-b) = a^2 - b^2}}}}}$

$\sf cos^2\theta = 1^2 - sin^2\theta$

${\bold{\blue{\boxed{\bf{sin^2\theta + cos^2\theta = 1 }}}}}$

${\bold{\blue{\boxed{\bf{sin^2 \theta - 1 = cos^2\theta }}}}}$

$\sf cos^2 \theta = cos^2 \theta$

${\bold{\red{\boxed{\bf{\dag cos^2 \theta = cos^2 \theta}}}}}$

${\bold{\pink{\boxed{\bf{\dag LHS = RHS}}}}}$

${\bold{\green{\boxed{\bf{\dag Hence Proved}}}}}$

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