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

Given equation is: tan A + sin A tan A - sin A = sec A + 1 sec A - 1 Taking the LHS of the equation, tan A + sin A tan A - sin A = sin A cos A + sin A sin A cos A - sin A = sin A + sin A cos A cos A ÷ sin A - sin A cos A cos A = sin A + sin A cos A cos A × cos A sin A - sin A cos A = sin A + sin A cos A sin A - sin A cos A = sin A 1 + cos A sin A 1 - cos A = 1 + cos A 1 - cos A Dividing all the terms in numerator and denominator by cos A , = 1 cos A + 1 1 cos A - 1 = sec A + 1 sec A - 1 = RHS Hence Proved.

If 7 sin 2 θ + 3 cos 2 θ = 4 and θ is acute, the show that cot θ = 3 .

Given, 7 sin 2 θ + 3 cos 2 θ = 4 and θ is acute. Now, 7 sin 2 θ + 3 cos 2 θ = 4 ⇒ 4 sin 2 θ + 3 cos 2 θ + 3 sin 2 θ = 4 ⇒ 4 sin 2 θ + 3 cos 2 θ + sin 2 θ = 4 ⇒ 4 sin 2 θ + 3 × 1 = 4 ∵ cos 2 θ + sin 2 θ = 1 ⇒ 4 sin 2 θ = 4 - 3 ⇒ 4 sin 2 θ = 1 ⇒ sin 2 θ = 1 4 ⇒ sin θ = 1 4 ⇒ sin θ = 1 2 ____________ 1 Now, cos 2 θ + sin 2 θ = 1 ⇒ cos 2 θ + 1 2 2 = 1 [Using 1 ] ⇒ cos θ = 1 - 1 4 ⇒ cos θ = 3 2 ____________ 2 Thus, cot θ = cos θ sin θ = 3 2 ÷ 1 2 [Using 1 and 2 ] = 3 2 × 2 = 3 Therefore, cot θ = 3 Hence, proved.

Prove the identity 1 - tan θ 1 - c o t θ 2 = tan 2 θ

We need to prove 1 - tan θ 1 - c o t θ 2 = tan 2 θ We know that, c o t θ = 1 tan θ L.H.S 1 - tan θ 1 - c o t θ 2 = 1 - tan θ 1 - 1 tan θ 2 1 - tan θ tan θ - 1 tan θ 2 = tan θ 1 - tan θ tan θ - 1 2 tan θ 1 - tan θ - 1 - tan θ 2 = - tan θ 2 = tan 2 θ Hence, 1 - tan θ 1 - c o t θ 2 = tan 2 θ

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

By taking LHS, L H S = 1 + sin A 1 - sin A On rationalizing the LHS, 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 2 cos 2 A ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = 1 cos A + sin A cos A = sec A + tan A ∵ 1 cos θ = s e c θ , sin θ cos θ = tan θ = R H S Hence, proved.

If tan θ = a b , then show that a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 .

Given that, tan θ = a b . We have to prove that, a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 LHS = a sin θ - b cos θ a sin θ + b cos θ = a sin θ cos θ - b cos θ cos θ a sin θ cos θ + b cos θ cos θ [Divide both the numerator and the denominator by cos θ ] = a tan θ - b a tan θ + b = a × a b - b a × a b + b [Substitute the value of tan θ = a b that is given] = a 2 - b 2 a a 2 + b 2 b = a 2 - b 2 a 2 + b 2 = RHS [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.

In the figure, ∠ B = 90 ° , ∠ C = 44 ° Prove that tan 44 ° × tan 46 ° = 1

LHS = tan 44 ° × tan 46 ° = tan 44 ° × c o t ( 90 ° - 46 ° ) ∵ tan θ = c o t 90 - θ = tan 44 ° × c o t 44 ° = tan 44 ° × 1 tan 44 ° = 1 = RHS

Prove that 1 + tan 2 A 1 + c o t 2 A = tan 2 A

Given, 1 + tan 2 A 1 + c o t 2 A = tan 2 A ⇒ 1 + sin 2 A cos 2 A 1 + cos 2 A sin 2 A ⇒ cos 2 A + sin 2 A cos 2 A sin 2 A + cos 2 A sin 2 A ⇒ 1 cos 2 A 1 sin 2 A ∴ sin 2 A + cos 2 A = 1 ⇒ sin 2 A cos 2 A ⇒ tan 2 A Hence, proved.

In ∆ ABC , ∠ B = 90 ° and tanC = 5 12 . Find the value of cos C + sin C sin C .

Consider the given figure It is given that tanC = 5 12 We know that cot C = 1 tan C ⇒ cot C = 12 5 . . . . i . We need to find: cos C + sin C sin C cos C + sin C sin C = cos C sin C + 1 = cot C + 1 ∵ cot C = cos C sin C = 12 5 + 1 [Using i ] = 17 5 Hence, the required value is 17 5 .

Prove that: sin A 1 + cot A - cos A 1 + tan A = sin A - cos A

LHS = sin A 1 + cot A - cos A 1 + tan A = sin A 1 + cos A sin A - cos A 1 + sin A cos A = sin A sin A + cos A sin A - cos A cos A + sin A cos A = sin 2 A sin A + cos A - cos 2 A cos A + sin A = sin 2 A - cos 2 A sin A + cos A = sin A + cos A sin A - cos A sin A + cos A = sin A - cos A = RHS Hence, proved.