Prove that : In triangle A B C , the least values of Σ cosec A 2 and Σ sec 2 A 2 are 6 and 4 respectively.

For triangle A B C , cosec A 2 , cosec B 2 , cosec C 2 are positive numbers. So, using A . M ≥ G . M , i.e. cosec A 2 + cosec B 2 + cosec C 2 3 ≥ cosec A 2 cosec B 2 cosec C 2 1 3 ⇒ cosec A 2 + cosec B 2 + cosec C 2 ≥ 3 1 sin A 2 sin B 2 sin C 2 1 3 ∵ sin A 2 sin B 2 sin C 2 ≤ 1 8 ⇒ 1 sin A 2 sin B 2 sin C 2 ≥ 8 So, ⇒ cosec A 2 + cosec B 2 + cosec C 2 ≥ 3 8 1 3 ⇒ cosec A 2 + cosec B 2 + cosec C 2 ≥ 6 Now, for real numbers x , y , z , we have x 2 + y 2 + z 2 - x y - y z - z x ≥ 0 ⇒ ∑ tan 2 A 2 - ∑ tan A 2 tan B 2 ≥ 0 ⇒ ∑ tan 2 A 2 ≥ ∑ tan A 2 tan B 2 ⇒ ∑ tan 2 A 2 ≥ 1 ∵ ∑ tan A 2 tan B 2 = 1 ⇒ ∑ sec 2 A 2 - 1 ≥ 1 ⇒ ∑ sec 2 A 2 ≥ 4 So, least values of Σ cosec A 2 and Σ sec 2 A 2 are 6 and 4 respectively.

Prove that: cosec x · cosec 2 x · sin 4 x · cos 6 x · cosec 10 x = cos 3 x sin 2 x sin 4 x + cos 5 x sin 4 x sin 6 x + cos 7 x sin 6 x sin 8 x + cos 9 x sin 8 x sin 10 x .

Let y 1 = cos 3 x sin 2 x sin 4 x = 2 sin x cos 3 x 2 sin x sin 2 x sin 4 x = sin 4 x - sin 2 x 2 sin x sin 2 x sin 4 x ∵ 2 cos A sin B = sin A + B - sin A - B = cosec x 2 ( cosec 2 x - cosec 4 x ) Similarly, y 2 = cos 5 x sin 4 x sin 6 x = 2 sin x cos 5 x 2 sin x sin 4 x sin 6 x = cosec x 2 [ cosec 4 x - cosec 6 x ] , y 3 = cos 7 x sin 6 x sin 8 x = 2 sin x cos 7 x 2 sin x sin 6 x sin 8 x = cosec x 2 [ cosec 6 x - cosec 8 x ] , y 4 = cos 9 x sin 8 x sin 10 x = 2 sin x cos 9 x 2 sin x sin 8 x sin 10 x = cosec x 2 [ cosec 8 x - cosec 10 x ] ∴ y 1 + y 2 + y 3 + y 4 = cosec x 2 [ cosec 2 x - cosec 10 x ] = cosec x [ sin 10 x - sin 2 x ] 2 sin 2 x sin 10 x = cosec x 2 cos 10 x + 2 x 2 sin 10 x - 2 x 2 2 sin 2 x sin 10 x = cosec x 2 cos 6 x sin 4 x 2 sin 2 x sin 10 x = cosec x · cosec 2 x · sin 4 x · cos 6 x · cosec 10 x

If tan α = p q where α = 6 β , α being acute angle, prove that; 1 2 ( p cosec 2 β - q sec 2 β ) = p 2 + q 2 .

L . H . S , 1 2 p cosec 2 β - q sec 2 β = 1 2 p sin 2 β - q cos 2 β Multiply and divide by q = q 2 p q sin 2 β - 1 cos 2 β = q 2 tan 6 β sin 2 β - 1 cos 2 β ∵ p q = tan 6 β = q 2 sin 6 β cos 2 β - cos 6 β sin 2 β sin 2 β cos 2 β cos 6 β ⇒ q 1 sin 4 β sin 4 β cos 6 β ∵ 2 sin θ cos θ = sin 2 θ & sin A cos B - cos A sin B = sin A - B = q cos 6 β = q q p 2 + q 2 = p 2 + q 2 = R . H . S

If tan π 4 + y 2 = tan 3 π 4 + x 2 , prove that sin y = sin x 3 + sin 2 x 1 + 3 sin 2 x

Given: tan π 4 + y 2 = tan 3 π 4 + x 2 Now, we know that tan θ = 1 - cos 2 θ 1 + cos 2 θ Therefore, tan π 4 + y 2 = 1 - cos π 2 + y 1 + cos π 2 + y ⇒ tan π 4 + y 2 = 1 + sin y 1 - sin y . . . . i Similarly, tan π 4 + x 2 = 1 + sin x 1 - sin x . . . . ii Now, we have tan π 4 + y 2 = tan 3 π 4 + x 2 ⇒ 1 + sin y 1 - sin y = 1 + sin x 1 - sin x 3 [Using i & ii ] ⇒ sin y + 1 sin y - 1 = sin x + 1 sin x - 1 3 ⇒ sin y + 1 sin y - 1 = sin 3 x + 1 + 3 sin 2 x + 3 sin x sin 3 x - 1 - 3 sin 2 x + 3 sin x Applying Componendo and Dibidendo rule, we have 2 sin y 2 = 2 sin 3 x + 3 sin x 2 1 + 3 sin 2 x ⇒ sin y = sin x 3 + sin 2 x 1 + 3 sin 2 x

Prove that from the equality sin 4 α a + cos 4 α b = 1 a + b , a , b > 0 follows the relation sin 8 α a 3 + cos 8 α b 3 = 1 ( a + b ) 3 .

sin 4 α a + cos 4 α b = 1 a + b , a , b > 0 ⇒ ( a + b ) a sin 4 α + ( a + b ) b cos 4 α = 1 ⇒ sin 4 α + b a sin 4 α + a b cos 4 α + cos 4 α = 1 ⇒ sin 2 α + cos 2 α 2 - 2 sin 2 α cos 2 α + b a sin 4 α + a b cos 4 α = 1 ⇒ 1 - 2 sin 2 α cos 2 α + b a sin 4 α + a b cos 4 α = 1 ⇒ b a sin 2 α - a b cos 2 α 2 = 0 ⇒ b a sin 2 α = a b cos 2 α ⇒ tan 2 α = a b ⇒ tan α = a b ⇒ sin α = a a + b , cos α = b a + b Now, sin 8 α a 3 + cos 8 α b 3 = a 4 ( a + b ) 4 · 1 a 3 + b 4 ( a + b ) 4 · 1 b 3 = 1 ( a + b ) 3

Prove that 4 sin 27 ° = ( 5 + 5 ) 1 / 2 - ( 3 - 5 ) 1 / 2 .

We have to prove 4 sin 27 ° = 5 + 5 1 / 2 - 3 - 5 1 / 2 . Since 27 ° is not a standard angle, we have to prove using multiple angles or by converting into angles whose value we know. Now Let, θ = 27 ° then 2 θ = 54 ° = 90 ° - 36 ° ∴ sin 2 θ = sin 54 ° = cos 36 ° = 5 + 1 4 Now using sin 2 θ = 2 sin θ cos θ ⇒ 2 sin θ cos θ = 5 + 1 4 Now square both sides ⇒ 4 sin 2 θ 1 - sin 2 θ = 6 + 2 5 16 = 3 + 5 8 ∴ - sin 4 θ + sin 2 θ = 3 + 5 32 sin 4 θ - sin 2 θ + 3 + 5 32 = 0 Using quadratic formula, sin 2 θ = 1 ± 1 - 3 + 5 8 2 But sin 27 ° < sin 30 ° i.e. sin 27 ° < 1 2 . Hence, we reject value with + v e sign ∴ sin 2 θ = 1 - 5 - 5 8 2 ∴ 2 sin 2 θ = 1 - 5 - 5 8 Now, multiply both sides with 8 . ⇒ 16 sin 2 θ = 8 - 2 2 5 - 5 = 8 - 2 10 - 2 5 . . . i Now, in above expression put 8 = 5 + 5 + 3 - 5 and 10 - 2 5 = 5 + 5 3 - 5 to simplify. ⇒ 16 sin 2 θ = 5 + 5 + 3 - 5 - 2 5 + 5 3 - 5 = 5 + 5 1 / 2 - 3 - 5 1 / 2 2 or 4 sin θ = 5 + 5 1 / 2 - 3 - 5 1 / 2 Hence, Proved.

If A + B + C = π , prove that tan 2 A 2 + tan 2 B 2 + tan 2 C 2 ≥ 1 .

Since A + B + C = π ⇒ A 2 + B 2 = π 2 - C 2 ⇒ tan A 2 + B 2 = tan π 2 - C 2 ⇒ tan A 2 + B 2 = 1 tan C 2 ⇒ tan A 2 + tan B 2 1 - tan A 2 tan B 2 = 1 tan C 2 ⇒ tan A 2 tan C 2 + tan B 2 tan C 2 = 1 - tan A 2 tan B 2 ⇒ tan A 2 tan C 2 + tan B 2 tan C 2 + tan A 2 tan B 2 = 1 Let x = tan A 2 , y = tan B 2 , z = tan C 2 i.e. x y + y z + z x = 1 . . ( 1 ) Since x , y , z are real numbers, so x 2 + y 2 + z 2 - x y - y z - z x ≥ 0 ⇒ x 2 + y 2 + z 2 ≥ x y + y z + z x ⇒ tan 2 A 2 + tan 2 B 2 + tan 2 C 2 ≥ 1 Hence proved.

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 12 - Trigonometric Ratios, Functions and Identities>EXERCISE-4>Q 21

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Prove that :

In an acute angled triangle $A B C$ , the least values of $Σ secA$ and $Σ \tan^{2} A$ are $6$ and $9$ respectively.

Important Questions on Trigonometric Ratios, Functions and Identities

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 12 - Trigonometric Ratios, Functions and Identities>EXERCISE-4>Q 22

Prove that :

In triangle $A B C$ , the least values of $Σ cosec (\frac{A}{2})$ and $Σ \sec^{2} (\frac{A}{2})$ are $6$ and $4$ respectively.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 12 - Trigonometric Ratios, Functions and Identities>EXERCISE-4>Q 23

Prove that: $\frac{\sec^{2} A + \sec^{2} B + \sec^{2} C}{3} \geq (\sec A \sec B \sec C)^{\frac{2}{3}}$ $= \frac{\cos 3 x}{\sin 2 x \sin 4 x} + \frac{\cos 5 x}{\sin 4 x \sin 6 x} + \frac{\cos 7 x}{\sin 6 x \sin 8 x} + \frac{\cos 9 x}{\sin 8 x \sin 10 x}$ .