Trigonometric Ratios & Identity Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

BC = 15 cm and sin B = $4/5$

sin B = ${AC}/{AB} = 4/5$

then BC = 3m

But, BC = 15 (given)

then AC = 4 × 5 = 20

AB = 5 × 5 = 25.

Hence, the value of AB is 25 cm.

Answer: (b)

tan θ + sec θ = 2 ......(i)

As we know

⇒ $sec^2 θ - tan^2 θ$ = 1

⇒ (sec θ - tan θ)(sec θ + tan θ) = 1

⇒ sec θ - tan θ = $1/2$ ....(ii)

equation (i) - eq (ii)-

2 tan θ = 2 - 1/2

2 tan θ = $3/2 ⇒ tan θ = 3/4$

So, option (b) is correct.

Answer: (b)

sec x cosec x = 2

This value is possible is x = 45°

$tan^n x + cot^n x = (\text"tan x")^n + (\text"cot x")^n$

= $(tan 45°)^n + (cot 45°)^n = (1)^n + (1)^n = 2$

Answer: (b)

$cosec^4 α - cot^4$ α = 17 (Given)

⇒ $(cosec^2 α - cot^2 α) (cosec^2 α + cot^2 α)$ = 17

⇒ 1.$({1 + cos^2 α}/{sin^2 α})$ = 17

⇒ 2 - $sin^2 α = 17 sin^2 α$

⇒ $18 sin^2 α = 2 ⇒ sin^2 α = 1/9$

∴ sin α = $1/3$ (since, α lie in first quadrant)

Answer: (c)

Statement 1

1 = ${√{\text"1 - sin n"}}/{√{\text"1 + sin x"}} = √{{(\text"1 - sin x")(\text"1 - sin x")}/{(\text"1 - sin x")(\text"1 + sin x")}}$

= ${\text"1 - sin x"}/{√{1 - sin^2 x}} = {\text"1 - sin x"}/{√{cos^2 x}} = {\text"1 - sin x"}/{\text"cos x"}$

P = q

r = ${\text"cos x"}/{\text"1 + sin x"} = {\text"cos x" (\text"1 - sin x")}/{(\text"1 + sin x")(\text"1 - sin x")} = {\text"cos x" (\text"1 - sin x")}/{1 - sin^2 x}$

= ${\text"cos x" (\text"1 - sin x")}/{cos^2 x} = {\text"1 - sin x"}/{\text"cos x"}$

P = q = r

Now,

Statement 2

$p^2$ = qr

= ${\text"1 - sin x"}/{\text"cos x"} . {\text"cos x"}/{\text"1 + sin x"} = {\text"1 - sin x"}/{\text"1 + sin x"} = P^2$

So, Both are correct.