Prove that: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A .

To prove: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A Taking L . H . S . sin A - 2 sin 3 A 2 cos 3 A - cos A = sin A 1 - 2 sin 2 A cos A 2 cos 2 A - 1 = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A + cos 2 A [Since sin 2 A + cos 2 A = 1 ] = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A - cos 2 A = tan A cos 2 A - sin 2 A cos 2 A - sin 2 A = tan A = R . H . S . . [proved]

Prove that: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A .

To prove: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A Taking L . H . S . sin A - 2 sin 3 A 2 cos 3 A - cos A = sin A 1 - 2 sin 2 A cos A 2 cos 2 A - 1 = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A + cos 2 A [Since sin 2 A + cos 2 A = 1 ] = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A - cos 2 A = tan A cos 2 A - sin 2 A cos 2 A - sin 2 A = tan A = R . H . S . . [proved]

If 5 cos θ + 12 sin θ = 13 , then show that tan θ = 12 5 .

Given: 5 cos θ + 12 sin θ = 13 Divide both sides of the equation by cos θ . ⇒ 5 cos θ cos θ + 12 sin θ cos θ = 13 cos θ We know that, tan θ = sin θ cos θ ; sec θ = 1 cos θ ⇒ 5 + 12 tan θ = 13 sec θ ⇒ 5 + 12 tan θ 2 = 13 sec θ 2 ⇒ 5 2 + 12 tan θ 2 + 2 × 5 × 12 tan θ = 13 sec 2 ⇒ 25 + 144 tan 2 θ + 120 tan θ = 169 sec 2 θ ⇒ 25 + 144 tan 2 θ + 120 tan θ = 169 1 + tan 2 θ ⇒ 25 + 144 tan 2 θ + 120 tan θ = 169 + 169 tan 2 θ ⇒ 25 + 144 tan 2 θ + 120 tan θ - 169 - 169 tan 2 θ = 0 ⇒ - 25 tan 2 θ + 120 tan θ - 144 = 0 ⇒ 25 tan 2 θ - 120 tan θ + 144 = 0 ⇒ 5 2 tan 2 θ - 120 tan θ + 12 2 = 0 ⇒ 5 tan θ - 12 2 = 0 ⇒ 5 tan θ - 12 = 0 ⇒ tan θ = 12 5 Hence, proved.

Prove that: tan 2 A - tan 2 B = sin 2 A - sin 2 B cos 2 A cos 2 B

We have to find the value of tan 2 A - tan 2 B = sin 2 A - sin 2 B cos 2 A cos 2 B . LHS = tan 2 A - tan 2 B = sin 2 A cos 2 A - sin 2 B cos 2 B = sin 2 A cos 2 B - sin 2 B cos 2 A cos 2 A cos 2 B = sin 2 A ( 1 - sin 2 B ) - sin 2 B ( 1 - sin 2 A ) cos 2 A cos 2 B = sin 2 A - sin 2 A sin 2 B - sin 2 B + sin 2 A sin 2 B cos 2 A cos 2 B = sin 2 A - sin 2 B cos 2 A cos 2 B = RHS Hence, proved.

Prove that sin θ - 2 sin 3 θ 2 cos 3 θ - cos θ = tan θ .

We have to prove that, sin θ - 2 sin 3 θ 2 cos 3 θ - cos θ = tan θ Proof: LHS = sin θ ( 1 - 2 sin 2 θ ) cos θ ( 2 cos 2 θ - 1 ) = sin θ ( cos 2 θ + sin 2 θ - 2 sin 2 θ ) cos θ ( 2 cos 2 θ - sin 2 θ - cos 2 θ ) [ ∵ sin 2 θ + cos 2 θ = 1 ] = sin θ ( cos 2 θ - sin 2 θ ) cos θ ( cos 2 θ - sin 2 θ ) = tan θ [ ∵ tan A = sin A cos A ] = RHS Hence, proved .

If sec θ + tan θ = x , then show that sin θ = x 2 - 1 x 2 + 1 .

Given: sec θ + tan θ = x ----- (i) We know that, sec 2 θ = 1 + tan 2 θ ⇒ sec 2 θ - tan 2 θ = 1 By using a 2 - b 2 = a + b a - b ⇒ sec θ + tan θ sec θ - tan θ = 1 ⇒ x sec θ - tan θ = 1 ⇒ sec θ - tan θ = 1 x ----- (ii) Adding ( i ) , ( i i ) ⇒ 2 sec θ = x + 1 x ⇒ 2 sec θ = x 2 + 1 x ⇒ sec θ = x 2 + 1 2 x By the definition; sec θ = 1 cos θ ⇒ cos θ = 2 x x 2 + 1 By using the identity: sin 2 θ + cos 2 θ = 1 ⇒ sin 2 θ = 1 - cos 2 θ ⇒ sin 2 θ = 1 - 2 x x 2 + 1 2 ⇒ sin 2 θ = x 2 + 1 2 - 2 x 2 x 2 + 1 2 ⇒ sin 2 θ = x 4 + 1 + 2 x 2 - 4 x 2 x 2 + 1 2 ⇒ sin 2 θ = x 4 + 1 - 2 x 2 x 2 + 1 2 = x 2 - 1 2 x 2 + 1 2 ⇒ sin θ = x 2 - 1 x 2 + 1 2 ∴ sin θ = x 2 - 1 x 2 + 1 Hence, proved.

Prove that sin A - cos A + 1 sin A + cos A - 1 = 1 sec A - tan A .

We have to prove sin A - cos A + 1 sin A + cos A - 1 = 1 sec A - tan A . L H S = sin A - cos A + 1 sin A + cos A - 1 On dividing by cos A in numerator and denominator, we get = sin A cos A - 1 + 1 cos A sin A cos A + 1 - 1 cos A = tan A - 1 + sec A tan A + 1 - sec A ∵ sin A cos A = tan A and 1 cos A = sec A = tan A + sec A - 1 tan A - sec A + 1 Now, on multiplying by tan A - sec A in numerator and denominator, we get = tan A + sec A - 1 tan A - sec A + 1 × tan A - sec A tan A - sec A = tan A + sec A tan A - sec A - tan A - sec A tan A - sec A + 1 tan A - sec A = tan 2 A - sec 2 A - tan A - sec A tan A - sec A + 1 tan A - sec A = - 1 - tan A - sec A tan A - sec A + 1 tan A - sec A ∵ sec 2 A - tan 2 A = 1 = - 1 1 + tan A - sec A tan A - sec A + 1 tan A - sec A = - 1 tan A - sec A = 1 sec A - tan A = R H S Hence, proved.

If tan 2 A = cot A - 18 ° , where 2 A is an acute angle, find the value of A

Given that, 2 A is an acute angle. ⇒ 2 A < 90 ° So, using trigonometric ratio of complementary angles, we can say that cot 90 ° - 2 A = tan 2 A ∵ cot 90 ° - θ = tan θ Given that, tan 2 A = cot A - 18 ° Now, replace tan 2 A by cot 90 ° - 2 A in the given question. ⇒ cot 90 ° - 2 A = cot A - 18 ° Now, we can compare the degrees from above, we get 90 ° - 2 A = A - 18 ° ⇒ 2 A + A = 90 ° + 18 ° ⇒ 3 A = 108 ° ⇒ A = 108 3 = 36 ° Thus, the value of A is 36 ∘

Prove that sin θ + cosecθ 2 + cos θ + sec θ 2 = 7 + tan 2 θ + cot 2 θ

Given, sin θ + cosecθ 2 + cos θ + sec θ 2 = 7 + tan 2 θ + cot 2 θ LSH: sin θ + cosec θ 2 + cos θ + sec θ 2 Expanding the binomials we get, sin 2 θ + 2 sin θ cosec θ + cosec 2 θ + cos 2 θ + 2 cos θ sec θ + sec 2 θ = sin 2 θ + cos 2 θ + 2 sin θ 1 sin θ + cosec 2 θ + sec 2 θ + 2 cos θ 1 cos θ [ cosec θ = 1 sin θ , sec θ = 1 cos θ ] = 1 + 2 + 2 + cosec 2 θ + sec 2 θ = 5 + 1 + cot 2 θ + 1 + tan 2 θ [ sec 2 θ = 1 + tan 2 θ ; cosec 2 θ = 1 + cot 2 θ ] = 7 + tan 2 θ + cot 2 θ = RHS Hence proved

If 4 tanθ = 3 , evaluate 4 sinθ - cosθ + 1 4 sinθ + cosθ - 1 .

Given that, tan θ = 3 4 ∴ tan 2 θ = 9 16 We know that, sec 2 θ = 1 + tan 2 θ . ∴ sec 2 θ = 1 + 9 16 = 25 16 ∴ cos 2 θ = 16 25 ∴ cos θ = 4 5 We know that, sin 2 θ = 1 - cos 2 θ ∴ sin 2 θ = 1 - 16 25 = 9 25 ⇒ sin θ = 3 5 Now, 4 sin θ - cos θ + 1 4 sin θ + cos θ - 1 = 4 × 3 5 - 4 5 + 1 4 × 3 5 + 4 5 - 1 = 12 - 4 + 5 12 + 4 - 5 = 13 11 .