Prove that (1+ cot a+ cosec a)(1+ cot a-cosec a)= cot a/2 -tan a/2

Open in App

Solution

$LHS = (1 + \cot a + cosec a) (1 + \cot a - cosec a) = {(1 + \cot a)}^{2} - {(cosec a)}^{2} = 1 + \cot^{2} a + 2 \cot a - {cosec}^{2} a = 1 + 2 \cot a - ({cosec}^{2} a - \cot^{2} a) = 1 + 2 \cot a - 1 = 2 \cot a RHS = \cot (\frac{a}{2}) - \tan (\frac{a}{2}) = \frac{\cos (a / 2)}{\sin (a / 2)} - \frac{\sin (a / 2)}{\cos (a / 2)} = \frac{\cos^{2} (a / 2) - \sin^{2} (a / 2)}{\sin (a / 2) . \cos (a / 2)} = \frac{\cos a}{\sin (a / 2) . \cos (a / 2)} = \frac{2 \cos a}{2 \sin (a / 2) . \cos (a / 2)} = \frac{2 \cos a}{\sin a} = 2 \cot a So, LHS = RHS$

Suggest Corrections

Similar questions

\frac{1 + c o s e c A + cot A}{1 + c o s e c A - cot A} = \frac{c o s e c A + cot A - 1}{cot A - c o s e c A + 1}

Prove that :

tanA/1-cotA + cotA/1-tanA = secAcosecA+1

(t a n A + c o s e c B)^{2} - (c o t B - s e c A)^{2} = 2 t a n A c o t B (c o s e c A + s e c B)

\cot^{2} A - \tan^{2} A = 4 \cot 2 A cosec 2 A

(1 + c o t A)^{2} + (1 - c o t A)^{2} = 2 c o s e c^{2} A

Join BYJU'S Learning Program