Question

Prove the following trigonometric identities:
cosecA/cosecA-1+cosecA/cosecA+1=2sec²A

Answer 1

• cosecA/cosecA-1+cosecA/cosecA+1=2sec²A
• L.H.S. = cosecA [1/cosecA-1 + 1/cosecA+1]
• =cosecA [2cosecA/(cosec²A-1)]
• using the identity, 1+cot²A=cosec²A
• =cosecA [2cosecA/cot²A]
• =2cosec²A/cot²A
• put cosecA=1/sinA and cotA=cosA/sinA we get,
• =2[(1/sin²A)×(sin²A/cos²A)]
• sin²A gets cancelled and we get (1/cos²A)=sec²A
• L.H.S.= 2sec²A=R.H.S.

Answer 2

The trigonometric identity, $\dfrac{\csc A}{\csc A-1} -\dfrac{\csc A}{\csc A+1} =\sec^2 A$, proved.

Step-by-step explanation:

To prove that, the trigonometric identity:

$\dfrac{\csc A}{\csc A-1} -\dfrac{\csc A}{\csc A+1} =\sec^2 A$

L.H.S. = $\dfrac{\csc A}{\csc A-1} -\dfrac{\csc A}{\csc A+1}$

= $\dfrac{\csc A(\csc A+1)+\csc A(\csc A-1)}{(\csc A-1)(\csc A+1)}$

= $\dfrac{\csc^2 A+\csc A+\csc^2 A-\csc A}{(\csc A-1)(\csc A+1)}$

Using the algebraic identity,

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

= $\dfrac{\csc^2 A+\csc^2 A}{\csc^2 A-1^2}$

= $\dfrac{2\csc^2 A}{\csc^2 A-1}$

Using the trigonometric identity,

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

⇒ $\cot^2 A=\csc^2 A-1$

= $\dfrac{2\csc^2 A}{\cot^2 A}$

Using the trigonometric identity,

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

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

= $\dfrac{2}{\cos^2 A}}$

= $\sec^2 A$

= R.H.S., proved.

Thus, the trigonometric identity, $\dfrac{\csc A}{\csc A-1} -\dfrac{\csc A}{\csc A+1} =\sec^2 A$, proved.