Question

Prove that: (1 + cot A - cosec A) (1 + tan A + sec A)=2
​

Answer 1

$\sf\small\underline\purple{Solution:–}$

LHS = (1+ cot a – cosec a)(1 + tan a – sec a)

LHS = (1 + cos a/sin a – 1/sin a) (1 + sin a/cos a – 1/cos a)

LHS = (sin a +cos a – 1/sin a) (cos a + sin -1/cos a)

LHS = [(sin a+cos a) – (1)/sin a] [(cos a + sin a) – (1)/cos a]

LHS = (sin a + cos a)2 – (1)2)/sin a cos a

LHS = sin 2a + cos 2a + 2sin acos a -1/sin a cos a

Since sin 2a + cos 2a = 1

LHS = 1+ 2 sin a cos a – 1/sin acos a

LHS = 2 sin a cos a/sin a cos a

LHS = 2

LHS = RHS = 2

Hence , proved !

Answer 2

Just open the bracket and keep taking LCM ull get the ans

LHS = (1+ cot a – cosec a)(1 + tan a – sec a)

LHS = (1 + cos a/sin a – 1/sin a) (1 + sin a/cos a – 1/cos a)

LHS = (sin a +cos a – 1/sin a) (cos a + sin -1/cos a)

LHS = [(sin a+cos a) – (1)/sin a] [(cos a + sin a) – (1)/cos a]

LHS = (sin a + cos a)2 – (1)2)/sin a cos a

LHS = sin 2a + cos 2a + 2sin acos a -1/sin a cos a

Since sin 2a + cos 2a = 1

LHS = 1+ 2 sin a cos a – 1/sin acos a

LHS = 2 sin a cos a/sin a cos a

LHS = 2

LHS = RHS = 2

hope it helps

pls mark me as brainliest