Prove that:-
(Whole root)CosecA+1
--------------- = SecA + TanA
CosecA-1

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Prove that:-
(Whole root)CosecA+1
--------------- = SecA + TanA
CosecA-1

Ankush Jain · Accepted Answer

To prove: $\sqrt{\frac{\cos e c A + 1}{\cos e c A - 1}} = \sec A + \tan A$ Proof : LHS = $\sqrt{\frac{\frac{1}{\sin A} + 1}{\frac{1}{\sin A} - 1}} = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$ On rationalizing the denominator, we get LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A} \times (\frac{1 + \sin A}{1 + \sin A})} = \sqrt{\frac{{(1 + \sin A)}^{2}}{1 - \sin^{2} A}} = \sqrt{\frac{{(1 + \sin A)}^{2}}{\cos^{2} A}}$ = $\frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A}$ = sec A + tan A = RHS [Hence Proved]

Prove that:- (Whole root)CosecA+1 --------------- = SecA + TanA CosecA-1

Prove that:-
(Whole root)CosecA+1
--------------- = SecA + TanA
CosecA-1