Math, asked by shahsanjay2630, 9 months ago

Prove the following trigonometric identities:
cos/cosecθ+1+cosθ/cosecθ-1=2tanθ

Answers

Answered by harendrachoubay
1

\dfrac{\cos \theta}{\csc \theta+1} +\dfrac{\cos \theta}{\csc \theta-1}=2\tan \theta, proved.

Step-by-step explanation:

To prove that:

\dfrac{\cos \theta}{\csc \theta+1} +\dfrac{\cos \theta}{\csc \theta-1}=2\tan \theta.

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

To take LCM of denominator part, we get

= \dfrac{\cos \theta(\csc \theta-1)+\cos \theta(\csc \theta+1)}{(\csc \theta+1)(\csc \theta-1)}

Using the algebraic identity,

a^{2} -b^{2} = (a + b)(a - b)

= \dfrac{\cos \theta\csc \theta-\cos \theta+\cos \theta\csc \theta+\cos \theta}{\csc^2 \theta-1^2}

= \dfrac{\cos \theta\csc \theta+\cos \theta\csc \theta}{\csc^2 \theta-1}

= \dfrac{2\cos \theta\csc \theta}{\csc^2 \theta-1}

Using the trigonometric identity,

\csc^2 A - \cot^2 A= 1

\cot^2 A = \csc^2 A -1

= \dfrac{2\cos \theta\csc \theta}{\cot^2 \theta}

= \dfrac{2\cos \theta\csc \theta}{\cot^2 \theta}

Using the trigonometric identity,

\csc A =\dfrac{1}{\sin A} and \cot A =\dfrac{\cos A }{\sin A}

= \dfrac{2\cos \theta\dfrac{1}{\sin A}}{\dfrac{\cos^2 A }{\sin^2 A}}

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

Using the trigonometric identity,

\tan A =\dfrac{\sin A }{\cos A}

= 2\tan \theta

= R.H.S., proved.

\dfrac{\cos \theta}{\csc \theta+1} +\dfrac{\cos \theta}{\csc \theta-1}=2\tan \theta, proved.

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