Question

prove that 1+cos theta /sin theta -sin theta /1+cos theta =2 cot theta ​

Answer 1

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Answer 2

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

Step-by-step explanation:

To prove that: $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta.$

L.H.S. = $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta}$

To take LCM of denominator part, we get

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

Using the algebraic identity,

$(a+b)^{2}$ = $a^{2}$ + 2ab + $b^{2}$

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

Using the trigonometric identity,

$\sin^2 A+\cos^2 A$ = 1

⇒ $\cos^2 A$ = 1 - $\sin^2 A$

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

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

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

Taking $2\cos \theta$ numerator as common, we get

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

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

Using the trigonometric identity,

$\cot A$ = $\dfrac{\cos A}{\sin A}$

= $2\cot \theta$

= R.H.S. proved.

Thus, $\dfrac{1+\cos \theta}{\sin \theta} -\dfrac{\sin \theta}{1+\cos \theta} =2\cot \theta$, proved.