Prove that 1 - sin A 1 + sin A = s e c A - t a n A

LHS: 1 - sin ⁡ A 1 + sin ⁡ A By rationalising the denominator we get, = 1 - sin A 1 + sin A × 1 - sin A 1 - sin A = 1 - sin ⁡ A 2 1 - sin 2 ⁡ A a + b a - b = a 2 - b 2 = 1 - sin ⁡ A cos 2 ⁡ A = 1 - sin ⁡ A cos ⁡ A = 1 cos ⁡ A - sin ⁡ A cos ⁡ A = sec ⁡ A - tan ⁡ A

Prove that 1 - sin A 1 + sin A = s e c A - t a n A

LHS: 1 - sin ⁡ A 1 + sin ⁡ A By rationalising the denominator we get, = 1 - sin A 1 + sin A × 1 - sin A 1 - sin A = 1 - sin ⁡ A 2 1 - sin 2 ⁡ A a + b a - b = a 2 - b 2 = 1 - sin ⁡ A cos 2 ⁡ A = 1 - sin ⁡ A cos ⁡ A = 1 cos ⁡ A - sin ⁡ A cos ⁡ A = sec ⁡ A - tan ⁡ A

Prove that: cos A cosec A − sin A sec A cos A + sin A = cosec A − sec A .

We have to prove that cos A cosec A − sin A sec A cos A + sin A = cosec A − sec A Solving LHS = cos A cosec A − sin A sec A cos A + sin A = cos A × 1 sin A − sin A × 1 cos A cos A + sin A = cos A sin A − sin A cos A cos A + sin A = cos 2 A - sin 2 A sin A · cos A · ( cos A + sin A ) ∵ a 2 - b 2 = ( a + b ) ( a - b ) = ( sin A + cos A ) ( cos A − sin A ) sin A cos A ( cos A + sin A ) = cos A − sin A sin A cos A = 1 sin A − 1 cos A = cosec A - sec A = RHS Hence proved.

Prove that : 1 sec α - tan α - 1 cos α = 1 cos α - 1 sec α + tan α

Prove that: 1 sec α - tan α - 1 cos α = 1 cos α - 1 sec α + tan α 1 sec α - tan α + 1 sec α + tan α = 1 cos α + 1 cos α 1 sec α - tan α + 1 sec α + tan α = 2 cos α L.H.S. = 1 sec α - tan α + 1 sec α + tan α = sec α + tan α + sec α - tan α sec 2 α - tan 2 α We know, sec 2 α - tan 2 α = 1 = 2 sec α 1 = 2 cos α = R.H.S. Hence proved

Prove that: cos 3 ⁡ A + sin 3 ⁡ A cos ⁡ A + sin ⁡ A + cos 3 ⁡ A - sin 3 ⁡ A cos ⁡ A - sin ⁡ A = 2

We have cos 3 ⁡ A + sin 3 ⁡ A cos ⁡ A + sin ⁡ A + cos 3 ⁡ A - sin 3 ⁡ A cos ⁡ A - sin ⁡ A = cos 3 ⁡ A + sin 3 ⁡ A cos ⁡ A - sin ⁡ A + ( cos 3 ⁡ A - sin 3 ⁡ A ) ( cos ⁡ A + sin ⁡ A ) cos 2 ⁡ A - sin 2 ⁡ A = cos 4 ⁡ A - cos 3 ⁡ A sin ⁡ A + sin 3 ⁡ A cos ⁡ A - sin 4 ⁡ A + cos 4 ⁡ A + cos 3 ⁡ A sin ⁡ A - sin 3 ⁡ A cos ⁡ A - sin 4 ⁡ A cos 2 ⁡ A - sin 2 ⁡ A = 2 ( cos 4 ⁡ A - sin 4 ⁡ A ) cos 2 ⁡ A - sin 2 ⁡ A = 2 ( cos 2 ⁡ A + sin 2 ⁡ A ) ( cos 2 ⁡ A - sin 2 ⁡ A ) cos 2 ⁡ A - sin 2 ⁡ A = 2 ( cos 2 ⁡ A + sin 2 ⁡ A ) = 2 ∵ sin 2 ⁡ A + cos 2 ⁡ A = 1 Hence, proved.

For all values of α , β , γ prove that, cos α + cos β + cos γ + cos α + β + γ = 4 cos α + β 2 cos β + γ 2 cos γ + α 2

LHS = cos α + cos β + cos γ + cos α + β + γ = 2 cos α + β 2 cos α - β 2 + 2 cos α + β + 2 γ 2 cos α + β 2 = 2 cos α + β 2 cos α - β 2 + cos α + β + 2 γ 2 = 2 cos α + β 2 2 cos α + γ 2 cos β + γ 2 = 4 cos α + β 2 cos β + γ 2 cos α + γ 2 = RHS

If x + y + z = π 2 show that, sin 2 x + sin 2 y + sin 2 z = 4 cos x cos y cos z .

We have, sin 2 x + sin 2 y + sin 2 z = 2 sin x + y cos x - y + sin 2 π 2 - x - y = 2 sin x + y cos x - y + sin π - 2 x - 2 y = 2 sin x + y cos x - y + sin 2 x + y = 2 sin x + y cos x - y + 2 sin x + y cos x + y = 2 sin x + y cos x - y + cos x + y = 2 sin π 2 - z 2 cos x cos y = 4 cos x cos y cos z

If A + B + C = 2 S then prove that- cos S - A + cos S - B + cos S - C + cos S = 4 cos A 2 cos B 2 cos C 2 .

If, A + B + C = 2 S then prove that, cos S - A + cos S - B + cos S - C + cos S = 4 cos A 2 cos B 2 cos C 2 We know that A + B + C = 180 0 (Sum of angles) S = 90 0 L.H.S. = cos S - A + cos S - B + cos S - C + cos S = cos 90 0 - A + cos 90 0 - B + cos 90 0 - C + cos 90 0 = sin A + sin B + sin C = 2 sin A + B 2 . cos A - B 2 + sin C A + B 2 = 90 0 - C 2 = 2 sin 90 0 - C 2 . cos A - B 2 + sin C = 2 cos C 2 . cos A - B 2 + 2 sin C 2 . cos C 2 = 2 cos C 2 cos A - B 2 + sin C 2 sin C 2 = cos A + B 2 = 2 cos C 2 cos A - B 2 + cos A + B 2 = 2 cos C 2 2 cos A + B + A - B 2 . cos A + B - A + B 2 = 2 cos C 2 2 cos A 2 . cos B 2 = 4 cos A 2 . cos B 2 . cos C 2 = R.H.S.