Question

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

Answer 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.