Question

cos e
sin 0
1-tan 0 1 - coto
+
= cos 0 + sin 6​

Answer 1

Answer:

$</p><p>{\green{\underline{\underline{\huge{\mathfrak\pink{your\:answer\:attached\:in\:the\:photo}}}}}}$

Answer 2

To prove:

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

Solution:

L.H.S

$\dfrac{\cos\theta}{1 - \tan\theta} + \dfrac{ \sin\theta}{1 - \cot\theta} \\ \\ = \dfrac{\cos\theta}{1 - \frac{ \sin\theta }{ \cos\theta} } + \dfrac{ \sin}{1 - \frac{\cos}{\sin\theta} } \\ \\ = \dfrac{\cos\theta}{ \frac{\cos\theta - \sin\theta}{ \cos\theta } } + \dfrac{ \sin\theta}{ \frac{\sin\theta - \cos\theta}{ \sin\theta}} \\ \\ = \frac{ { \cos }^{2} \theta}{ \cos\theta - \sin\theta } + \frac{ { \sin }^{2} \theta}{ \sin\theta - \cos\theta } \\ \\ = \frac{ { \cos}^{2}\theta }{ \cos\theta - \sin\theta } -\frac{ { \sin }^{2}\theta }{ \cos\theta - \sin\theta } \\ \\ = \frac{ { \cos}^{2}\theta - \sin{}^{2}\theta }{ \cos\theta - \sin\theta } \\ \\ = \frac{ ( { \cos\theta} + \sin \theta)({ \cos\theta} - \sin\theta)}{( \cos\theta - \sin\theta )} \\ \\ = \cos\theta + \sin\theta$

Hence, L.H.S = R.H.S [Proved]

Formula Used

a² - b² = (a + b)(a - b)

tan ∅ = sin ∅/cos ∅

cot ∅ = cos ∅/sin ∅