Question

prove that 1 + cos theta divided by sin theta minus sin theta divided by 1 + cos theta is equal to 2 cot​

Answer 1

Answer:

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Answer 2

LHS = RHS

Step-by-step explanation:

In this expression, we need to prove that,

$\dfrac{1+\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{1+\cos\theta}=2\cot\theta$

Taking LHS,

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

Using $(a+b)^2=a^2+b^2=2ab$

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

Since, $\sin^2\theta=1-\cos^2\theta$

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

Since, $\dfrac{\cos\theta}{\sin\theta}=\cot\theta$

$=\dfrac{2\cos\theta}{\sin\theta}\\\\=\cot\theta\\\\=RHS$

Hence, LHS = RHS

Learn more,

Trigonometry

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