Trigonometric Ratios & Identity Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

∵ 5 sin θ + 12 cos θ = 12

Now, squaring both sides, we get

$25 sin^2 θ + 144 cos^2 θ + \text"120 sin θ cos θ"$ = 169

⇒ 25$(1 - cos^2 θ) + 144(1 - sin^2 θ) + \text"120 sin θ cos θ"$ = 169

⇒ 25 - 25 $ cos^2 θ + 144 - 144 sin^2 θ + \text"120 sin θ cos θ"$ - 169

⇒ 25 $cos^2 θ + 144 sin^2 θ - \text"120 sin θ cos θ" = 169 - 169$

⇒ $(\text"5 cos θ - 12 sin θ")^2$ = 0

∴ 5 cos θ - 12 sin θ = 0

Answer: (a)

Let 4 $sin^2$ θ + 1 ≥ 4 sin θ

4 $sin^2$ θ - 4 sin θ + 1 ≥ 0

$(2 sin θ - 1)^2 ≥ 0$

sin θ ≥ $1/2$

Answer: (b)

Given, x ${cosec^2 30° sec^2 45°}/{8 cos^2 45° sin^2 60°} = tan^2 60° - tan^2 30°$

⇒ ${x × (2)^2 × (√2)^2}/{8 × (1/{√2})^2 × ({√3}/2)^2} = (√3)^2 - (1/{√3})^2$

⇒ ${x × 4 × 2 × 4}/{8 × 1/2 × 3} = 3 - 1/3$

⇒ ${8x}/3 = 8/3 ⇒$ x = 1

Answer: (d)

Let AB be the height, AC be the string and the angle made by string with the post be θ.

From figure cos θ = ${AB}/{AC} = h/{2h} = 1/2 = cos {π}/3$

θ = ${π}/3$

Answer: (c)

We have,

$sin^2x$ + sin x = 1...(1)

∴ sin x = 1 - $sin^2 x = cos^2 x$

On cubing equation (1), we get

$(sin^2 x + sin x)^3 = {1}^3$

$sin^6 x + sin^3 x + 3 sin^2 x. sin x (sin^2 \text"x + sin x")$ = 1

$sin^6x + sin^3 x + 3 sin^5x + 3 sin^4x$ = 1

∴ $cos^{12}x + 3 cos^{10}x + 3 cos^8 x + co^6 x$ = 1

Answer: (d)

sin 25° sin 35° sec 65° sec 55°

= sin 25° . sin 35° . $1/{cos 65°} . 1/{cos 55°}$

= sin 25° . sin 35° . $1/{cos (90 - 25°)} . 1/{cos (90 - 35°)}$

= sin 25° . sin 35° . $1/{sin 25°} . 1/{sin 35°}$ = 1