If sin ⁡ θ = 12 13 , find the value of s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ .

Given: sin ⁡ θ = 12 13 ……(1) We know that, sin 2 θ + cos 2 θ = 1 . 12 13 2 + cos 2 θ = 1 ⇒ cos 2 θ = 1 - 12 13 2 = 169 - 144 169 ⇒ cos 2 θ = 25 169 = 5 13 2 ⇒ cos θ = 5 13 tan θ = sin θ cos θ = 12 13 5 13 = 12 5 Substituting the value of sin ⁡ θ , cos ⁡ θ and tan θ from the previous step, we get, s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ = 12 13 2 - 5 13 2 2 × 12 13 × 5 13 × 1 12 5 2 = 144 169 - 25 169 2 × 12 × 5 13 × 13 × 25 144 = 144 - 25 169 120 169 × 25 144 = 119 169 × 169 120 × 25 144 = 119 1 × 1 24 × 5 144 = 119 × 5 24 × 144 = 595 3456 Therefore, the value of s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ is 595 3456 .

If cos ⁡ θ = 5 13 , find the value of s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ .

Given: cos ⁡ θ = 5 13 Let us assume that the length of Base side adjacent to given angle = 5 x and Hypotenuse = 13 x The triangle representing angle θ can be drawn as shown below: Applying Pythagoras theorem, we get, A C 2 = A B 2 + B C 2 ⇒ 13 x 2 = A B 2 + 5 x 2 ⇒ A B 2 = 13 x 2 - 5 x 2 ⇒ A B 2 = 169 x 2 - 25 x 2 ⇒ A B 2 = 144 x 2 ⇒ A B = 144 x 2 ∴ A B = 12 x Now, we know that, s i n θ = P e r p e n d i c u l a r s i d e o p p o s i t e t o ∠ θ H y p o t e n u s e ⇒ s i n θ = A B A C = 12 13 , and t a n θ = s i n θ c o s θ ⇒ t a n θ = 12 13 5 13 ⇒ t a n θ = 12 5 Substituting the values obtained in the given expression, s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ = 12 13 2 - 5 13 2 2 × 12 13 × 5 13 × 1 12 5 2 = 144 169 - 25 169 2 × 12 × 5 13 × 13 × 25 144 = 119 169 × 169 120 × 25 144 = 119 1 × 1 24 × 5 144 = 119 × 5 24 × 144 = 595 3456 Therefore, the value of s i n 2 θ - c o s 2 θ 2 s i n θ c o s θ × 1 t a n 2 θ is 595 3456 .

If sec ⁡ A = 5 4 , verify that 3 s i n A - 4 s i n 3 A 4 c o s 3 A - 3 c o s A = 3 t a n A - tan 3 A 1 - 3 t a n 2 A .

Given: sec ⁡ A = 5 4 cos ⁡ A = 1 sec ⁡ A = 4 5 Now, we know the following trigonometric identity cos 2 ⁡ A + sin 2 ⁡ A = 1 s i n 2 A = 1 - 4 5 2 = 1 - 16 25 = 9 25 s i n A = 9 25 = 3 5 Now, we know that t a n A = s i n A c o s A . We get, t a n A = 3 5 4 5 = 3 4 L.H.S = 3 s i n A - 4 s i n . 3 A 4 cos 3 ⁡ A - 3 c o s A = 3 3 5 - 4 3 5 3 4 4 5 3 - 3 4 5 = 9 5 - 4 125 27 4 64 125 - 12 5 = 9 5 - 108 125 256 125 - 12 5 = 225 125 - 108 125 256 125 - 300 125 = 117 125 - 44 125 = - 117 44 ⇒ 3 s i n A - 4 s i n 3 A 4 c o s 3 A - 3 c o s A = - 117 44 … … ( i ) R.H.S = 3 tan A - t a n 3 A 1 - 3 tan 2 A = 3 3 4 - 3 4 3 1 - 3 3 4 2 = 9 4 - 27 64 1 - 3 × 9 16 = 9 × 16 4 × 16 - 27 64 16 - 27 16 = 144 64 - 27 64 - 11 16 = 144 - 27 64 - 11 16 = 117 64 - 11 16 = 117 64 × 16 - 11 = - 117 × 1 4 × 11 ⇒ 3 tan A - tan 3 A 1 - 3 tan 2 A = - 117 44 … . . i i Comparing equation ( i ) and ( i i ) , we get, 3 s i n A - 4 s i n 3 A 4 c o s 3 A - 3 c o s A = 3 t a n A - t a n 3 A 1 - 3 t a n 2 A

If sin ⁡ θ = 3 4 , prove that c o s e c 2 θ - c o t 2 θ s e c 2 θ - 1 = 7 3 .

Given: sin ⁡ θ = 3 4 Let us assume the triangle with length of Perpendicular side = 3 x and Hypotenuse = 4 x Applying the Pythagoras Theorem, A C 2 = A B 2 + B C 2 ⇒ 4 x 2 = 3 x 2 + B C 2 ⇒ B C 2 = 4 x 2 - 3 x 2 ⇒ B C 2 = 16 x 2 - 9 x 2 ⇒ B C 2 = 7 x 2 ⇒ B C = 7 x Now, taking the L.H.S. of the given expression, cosec 2 θ - cot 2 θ sec 2 θ - 1 = 1 tan 2 θ ∵ 1 + cot 2 θ = cosec 2 θ ; 1 + tan 2 θ = sec 2 θ = 1 tanθ = cotθ = Base Perpendicular = 7 3 Hence, proved.