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

Answer: (c)

tan(x + 40)° tan (x + 20)° tan 3x° tan (70 - x)° tan (50 - x)° = 1

⇒ tan(x + 40) tan(x + 20) tan 3x cot [90 - (70 - x)] cot [90 - (50 - x)] = 1

⇒ tan(x + 40) tan (x + 20) tan 3x cot (x + 20) cot (x + 40) = 1

⇒ tan 3x = tan 45°

⇒ 3x = 45°

⇒ x = 15°

So, option (c) is correct.

Answer: (a)

Given,

In ΔABC,

tan B = $k/{√3 k}$

By Pythogaros theorem,

$AB^2 + AC^2 = BC^2$

⇒ $(√3 k)^2 + (1k)^2 = BC^2$

⇒ $BC^2 = 4k^2$

⇒ BC = 2k

Answer: (a)

tan1° tan2° tan3° tan4°.......tan 89°

tan1° tan2° ......tan45°......tan 89°

= 1

Answer: (c)

Given that, cos x = $5/{13} = {Base}/{Hypotenuse}$

P = $√{h^2 - b^2} = √{13^2 - 5^2}$

= $√{169 - 25} = √{144}$ = 12

∴ tan x - cot x = $p/b - b/p$

= ${12}/5 - 5/{12} = {144 - 25}/{60} = {119}/{6}$

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: (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 ≥ $1/2$