Question

5
Prove that sinO/1-cotO+cosO/1-tanO=sinO+cosO
​

Answer 1

$\maltese \;\underline{\textbf{\textsf{\orange{Solution \: :-}}}}$

$\longrightarrow \small\sf \dfrac{sin\theta}{1-cot\theta} + \dfrac{cos\theta}{1-tan\theta}$

◍ $\underline{\sf\small Substituting \: :- }$

• $\small\cot\theta = \dfrac{\cos\theta}{\sin\theta}$

• $\small\tan\theta = \dfrac{\sin\theta}{\cos\theta}$

$\displaystyle\longrightarrow \small \sf \dfrac{sin\theta}{1-\dfrac{cos\theta}{sin\theta}} + \dfrac{cos\theta}{1-\dfrac{sin\theta}{cos\theta}}$

$\displaystyle\longrightarrow \small \sf \dfrac{sin\theta}{\dfrac{sin\theta - cos\theta}{sin\theta}} + \dfrac{cos\theta}{\dfrac{cos\theta - sin\theta}{cos\theta}}$

$\longrightarrow \small \sf \dfrac{sin^2\theta}{sin\theta-cos\theta} + \dfrac{cos^2\theta}{cos\theta - sin\theta}$

$\longrightarrow \small \sf \dfrac{sin^2\theta}{sin\theta-cos\theta} + \dfrac{cos^2\theta}{-(sin\theta - cos\theta)}$

$\longrightarrow \small \sf \dfrac{sin^2\theta}{sin\theta-cos\theta} - \dfrac{cos^2\theta}{cos\theta - sin\theta}$

$\longrightarrow \small \sf \dfrac{sin^2\theta - cos^2\theta}{cos\theta - sin\theta}$

◍ $\underline{\small\sf Using\: :-}$

• $\small\rm (a^2 - b^2) = (a+b)(a-b)$

$\longrightarrow \small \sf \dfrac{(sin\theta + cos\theta)\cancel{(sin\theta-cos\theta)}}{\cancel{cos\theta - sin\theta}}$

$\longrightarrow \small \underline{\boxed{\bf sin\boldsymbol\theta +cos\boldsymbol\theta}}$

Answer 2

Step-by-step explanation:

$\rm \large \underline{† \: \: Question \: \: †}$

$\bf \: Prove \: \: that \: \: \frac{sin \: \theta}{1 - \: \: cot \: \theta} + \frac{ cos \: \theta }{1 - \: tan \: \theta} = \: sin \: \theta \: + \: cos \: \theta$

$\rm \large \underline{† \: Answer \: †}$

$\tt \large \implies{ \frac{sin \: \theta}{1 - \: \: cot \: \theta} + \frac{ cos \: \theta }{1 - \: tan \: \theta} }$

$\tt \large \implies{ \frac{sin \: \theta}{1 - \frac{cos \theta}{sin \theta} } + \frac{cos \: \theta}{1 - \frac{sin \theta}{cos \: \theta} } }$

$\tt \large \implies{ \frac{ {sin}^{2} \theta }{sin \theta - cos \theta} + \frac{ {cos}^{2} \theta}{cos \theta - sin \theta}}$

$\tt \large \implies{ \frac{ {sin}^{2} \theta }{sin \theta - cos \theta} + \frac{ {cos}^{2} \theta}{ - ( - cos \theta + sin \theta)}}$

$\tt \large \implies \frac{ {sin}^{2} \theta}{sin \theta \: - \: cos \: \theta} - \frac{ {cos}^{2} \theta}{sin \theta \: - \: cos \: \theta}$

$\tt \large \implies \frac{ {sin}^{2} \theta - {cos}^{2} \theta }{sin \theta - cos \theta}$

(a²-b²) = (a+b)(a-b)

$\tt \large \implies \frac{(sin \theta \: + \: cos \theta) \cancel{(sin \theta - cos \theta)}}{\cancel{(sin \theta - cos \theta)}}$

${\bf {\large{ \boxed{ \red{ \frak{ \implies{sin \theta \: + \: cos \theta}}}}}}}$

•°• LHS = RHS !!

Hence Proved !!!

#Sanjushri Reddy