Find the general solution of the equation sin 2 θ = 1 2 .

Given that, sin 2 θ = 1 2 ⇒ sin 2 θ = sin π 6 General solution of sin θ = sin α is α = n π + ( - 1 ) n θ , n ∈ I . General solution of sin 2 θ = sin π 6 is 2 θ = n π + ( - 1 ) n π 6 , n ∈ I i.e, θ = n π 2 + ( - 1 ) n π 12 , n ∈ I .

Find the general solution of the equation sin 2 θ = 1 2 .

Given that, sin 2 θ = 1 2 ⇒ sin 2 θ = sin π 6 General solution of sin θ = sin α is α = n π + ( - 1 ) n θ , n ∈ I . General solution of sin 2 θ = sin π 6 is 2 θ = n π + ( - 1 ) n π 6 , n ∈ I i.e, θ = n π 2 + ( - 1 ) n π 12 , n ∈ I .

Solve the equation tan - 1 1 - x 1 + x = 1 2 tan - 1 x , ( x > 0 ) .

Consider the given equation : tan - 1 1 - x 1 + x = 1 2 tan - 1 x ⇒ 2 tan - 1 1 - x 1 + x = tan - 1 x ⇒ tan - 1 2 1 - x 1 + x 1 - 1 - x 1 + x 2 = tan - 1 x . ∵ 2 tan - 1 A = tan - 1 2 A 1 - A 2 ⇒ tan - 1 2 1 - x 1 + x 1 + x 2 - 1 - x 2 1 + x 2 = tan - 1 x ⇒ tan - 1 2 1 - x 1 + x 1 + x 2 - 1 - x 2 = tan - 1 x ⇒ tan - 1 2 1 - x 2 1 + x 2 + 2 x - 1 + x 2 - 2 x = tan - 1 x ⇒ tan - 1 1 - x 2 2 x = tan - 1 x By comparing the angles, we get: 1 - x 2 2 x = x ⇒ 1 - x 2 = 2 x 2 ⇒ 1 - 3 x 2 = 0 ⇒ x = ± 1 3 Since, it is given that x > 0 So, - 1 3 is not possible value of x . Hence, x = 1 3

Show that tan - 1 1 5 + tan - 1 1 7 + tan - 1 1 3 + tan - 1 1 8 = π 4

Consider the given equation: tan - 1 1 5 + tan - 1 1 7 + tan - 1 1 3 + tan - 1 1 8 = π 4 We know that tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x , y > 0 and x y < 1 L . H . S . = tan - 1 1 5 + tan - 1 1 7 + tan - 1 1 3 + tan - 1 1 8 -----(i) = tan - 1 1 5 + 1 7 1 - 1 5 × 1 7 + tan - 1 1 3 + 1 8 1 - 1 3 × 1 8 = tan - 1 12 35 34 35 + tan - 1 11 23 = tan - 1 6 17 + tan - 1 11 23 = tan - 1 6 17 + 11 23 1 - 6 17 × 11 23 = tan - 1 325 391 325 391 = tan - 1 1 = tan - 1 tan π 4 = π 4 = R . H . S . Hence proved.

In △ A B C , prove that sin B - C 2 = b - c a cos A 2 .

Consider the given equation sin B - C 2 = b - c a cos A 2 . We know that by Sine rule ⇒ a sin A = b sin B = c sin C = k ∴ ⇒ a = k sin A , b = k sin B and c = k sin C Now take the right hand side of the given equation: R . H . S . = b - c a cos A 2 = k sin B - sin C k sin A cos A 2 = sin B - sin C sin A cos A 2 ∵ sin C - sin D = 2 cos C + D 2 sin C - D 2 , sin 2 A = 2 sin A cos A ⇒ R . H . S . = 2 cos B + C 2 sin B - C 2 2 sin A 2 cos A 2 cos A 2 = cos π 2 - A 2 sin B - C 2 sin A 2 ∵ In △ ABC , A + B + C = π = sin A 2 sin B - C 2 sin A 2 = sin B - C 2 ⇒ R . H . S . = L . H . S . Hence proved.

In ∆ A B C , prove that: a c cos B - b c cos A = a 2 - b 2 .

To prove: a c cos B - b c cos A = a 2 - b 2 . Now, LHS = a c cos B - b c cos A = a c a 2 + c 2 - b 2 2 a c - b c b 2 + c 2 - a 2 2 b c [As, cos B = a 2 + c 2 - b 2 2 a c & cos A = b 2 + c 2 - a 2 2 b c ] = a 2 + c 2 - b 2 2 - b 2 + c 2 - a 2 2 = a 2 + c 2 - b 2 - b 2 - c 2 + a 2 2 = 2 a 2 - b 2 2 = a 2 - b 2 = RHS Hence, proved.

If cos - 1 x + cos - 1 y + cos - 1 z = π , then prove that x 2 + y 2 + z 2 + 2 x y z = 1 .

Given, cos - 1 x + cos - 1 y + cos - 1 z = π cos - 1 x + cos - 1 y = π – cos - 1 z We know that cos - 1 x + cos - 1 y = cos - 1 [ x y – ( 1 – x 2 ) ( 1 – y 2 ) ] and cos - 1 - x = π - cos - 1 x ⇒ cos - 1 [ x y – ( 1 – x 2 ) ( 1 – y 2 ) ] = cos - 1 ( - z ) ⇒ x y – ( 1 – x 2 ) ( 1 – y 2 ) = ( - z ) ⇒ x y + z = ( 1 – x 2 ) ( 1 – y 2 ) Squaring on both sides, ⇒ ( x y + z ) 2 = [ ( 1 – x 2 ) ( 1 – y 2 ) ] 2 ⇒ x 2 y 2 + z 2 + 2 x y z = ( 1 – x 2 ) ( 1 – y 2 ) ⇒ x 2 y 2 + z 2 + 2 x y z = 1 – x 2 – y 2 + x 2 y 2 ⇒ z 2 + 2 x y z = 1 – x 2 – y 2 ⇒ x 2 + y 2 + z 2 + 2 x y z = 1 Hence, proved.