Question

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Answer 1

To Prove -

$\frac{ \sin( \theta) }{ \cot( \theta) + \csc( \theta) } = 2 + \frac{ \sin( \theta) }{ \cot( \theta) - \csc( \theta) }$

Solution:-

Taking LHS -

$\sf{\implies \frac{ \sin( \theta) }{ \cot( \theta) + \csc( \theta) } }$

Converting cot and cosec in the form of sin and Cos.

$\sf{\implies \frac{ \sin( \theta) }{ \frac{ \cos( \theta) }{ \sin( \theta) } + \frac{1}{ \sin( \theta) } } }$

$\sf{\implies \frac{ \sin(\theta) }{ \frac{ \cos(\theta) + 1}{ \sin(\theta) } }}$

$\sf{\implies \: \frac{ \sin(\theta) \times \sin(\theta) }{ \cos(\theta) + 1 }}$

Sin²a = 1 - cos²a .

$\sf{\implies \: \frac{ 1 - {\cos}^{2} \theta }{ \cos(\theta) + 1 } }$

a² - b² = (a+b)(a-b)

$\sf{ \implies \: \frac{(1 - \cos(\theta))(1 + \cos(\theta)) }{1 + \cos(\theta) } }$

$\sf{ \implies \: \frac{ (1 - \cos(\theta)) \cancel{(1 + \cos(\theta)) } }{ \cancel{1 + \cos(\theta) }} }$

→ 1 - cos theta = LHS .

Now taking RHS .

$\sf{\implies 2 + \frac{ \sin( \theta) }{ \cot( \theta) - \csc( \theta) }}$

$\sf{\implies 2 + \frac{ \sin( \theta) }{ \frac{ \cos( \theta) }{ \sin( \theta) } - \frac{1}{ \sin( \theta) } } }$

$\sf{\implies 2 + \frac{ \sin(\theta) }{ \frac{ \cos(\theta) - 1}{ \sin(\theta) } }}$

$\sf{\implies \: 2 + \frac{ \sin(\theta) \times \sin(\theta) }{ \cos(\theta) - 1 }}$

Sin²a = 1 - cos²a .

$\sf{\implies \: 2+ \frac{ 1 - {\cos}^{2} \theta }{ \cos(\theta) - 1 } }$

a² - b² = (a+b)(a-b)

$\sf{ \implies \: 2+ \frac{(1 - \cos(\theta))(1 + \cos(\theta)) }{- ( \cos(\theta) - 1) } }$

$\sf{\implies \: \frac{ - 2 + 1 + \cos \theta }{-1} }$

$\sf{\implies \: \frac{ - 1 + \cos \theta }{-1} }$

→ 1 - Cos theta = RHS

LHS = RHS

Hence proved .

Answer 2

Question :

Prove :

$\sf\frac{\sin\theta}{\cot\theta+\cot\theta}=2+\frac{\sin\theta}{\cot\theta-\cosec\theta}$

Solution :

LHS $=\sf\frac{\sin\theta}{\cot\theta+\cot\theta}$

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

$=\sf\frac{\sin^2\theta}{(1+\cos\theta)}$

$=\sf\frac{1-\cos^2\theta}{(1+\cos\theta)}$

$=\sf\frac{(1+\cos\theta)(1-\cos\theta)}{(1+\cos\theta)}$

$\sf=1-\cos\theta$....(1)

RHS $=\sf2+\frac{\sin\theta}{\cot\theta-\cot\theta}$

$=\sf2+\dfrac{\sin\theta}{\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}}$

$=\sf2+\frac{\sin^2\theta}{(\cos\theta-1)}$

$=\sf2+\frac{1-\cos^2\theta}{(\cos\theta-1)}$

$=\sf2+\frac{(1+\cos\theta)(1-\cos\theta)}{\cos\theta-1)}$

$=\sf2+\frac{(1+\cos\theta)(1-\cos\theta)}{(-1)(1-\cos\theta)}$

$\sf=2-(1+\cos\theta)$...(2)

LHS = RHS

⇒1-cosθ=2-(1+cosθ)

⇒(1-cosθ)+(1-cosθ)=2

⇒2=2

Hence ,Proved