Trigonometric Ratios And Identities Model Questions & Answers, Practice Test for ssc constables capfs nia ssf ar tier 1 2024

Answer: (d)

tan A + cot A = 4

⇒ Squaring both sides

$tan^2 A + cot^2 A$ + 2 = 16

$tan^2 A + cot^2 A$ = 14

Again, squaring both sides

$tan^4 A + cot^4 A + 2$ = 196

$tan^4 A + cot^4 A$ = 194

Answer; (c)

In the first statement, we are given that 45° < θ < 60°.

Therefore, if we consider θ < 60°, then let us suppose θ = 45°, so

$sec^2 θ + cosec^2 θ = sec^2 45° + cosec^2 45°$

= 2 + 2 - 4 = $α^2$ ⇒ α = 2 > 1 and if we consider θ > 45°,

then let us suppose θ = 60°, so

$sec^2 θ + cosec^2 θ = sec^2 60° + cosec^2 60°$

= $4 + 4/3 = {16}/3 = α^2$ ⇒ α = 2.31 > 1

Hence, statement 1 is true.

In the second statement, we are given that 0° > θ > 45°. Therefore, if we consider θ < 45°, then let us suppose θ = 0°, so

${\text"1 + cos θ"}/{\text"1 - cos θ"} = {\text"1 + cos 0°"}/{\text"1 - cos 0°"} = ∞ = x^2 ⇒ x = ∞ > 2$

and if we consider θ > 0°, then let us suppose θ = 45°,

so

${\text"1 + cos θ"}/{\text"1 - cos θ"}= = {\text"1 + cos 45°"}/{\text"1 - cos 45°"}$

= ${√2 + 1}/{√2 - 1} × {√2 + 1}/{√2 + 1} = {(√2 + 1)^2}/{2 - 1}$

= ${2 + 1 + 2 √2}/1 = 3 + 2 √2 = x^2$ ⇒ x = 2.414 > 2

Hence, statement 2 is true.

In the third statement, we are given that 0° < θ < 45°.

Therefore, if we consider θ < 45°, then let us suppose θ = 0°, so

${\text"cos θ"}/{\text"1 - tan θ"} + {\text"sin θ"}/{\text"1 - cot θ"} = {\text"cos 0°"}/{\text"1 - tan 0°"} = {\text"sin 0°"}/{\text"1 - cot 0°"}$

= $1/{1 - 0} + 0/{- ∞}$ = 1 + 0 = 1

which is not ≥ 2 and if we consider θ > 0°, then let us suppose θ = 45°, so

${\text"cos θ"}/{\text"1 - tan θ"} + {\text"sin θ"}/{\text"1 - cot θ"} = {cos 45°}/{1 - tan 45°} + {sin 45°}/{1 - cot 45°}$

= ${1/{√2}}/{1 - 1} + {1/{√2}}/{1 - 1}$ = ∞ ≥ 2.

Hence, statement 3 is false.

Hence, only two statements are correct.

Answer: (a)

Given that, ${\text"sin θ"}/{\text"cos θ"} + {\text"cos θ"}/{\text"sin θ"}$ = 2

∴ $sin^2 θ + cos^2 θ$ = 2 sin θ cos θ

⇒ sin 2 θ = 1 = sin 90°

⇒ 2 θ = 90° ⇒ θ = 45°

Answer: (d)

$cosec^2 68° + sec^2 56° - cot^2 34° - tan^2 22°$

$sec^2 22° - tan^2 22° + cosec^2 34° - cot^2 34°$

1 + 1 = 2

Answer: (b)

${(\text"1 + sec θ - tan θ) cos θ"}/{(\text"1 + sec θ + tan θ") (\text"1 - sin θ")}$

= ${(1 + 1/{\text"cos θ"} - {\text"sin θ"}/{\text"cos θ"}) \text"cos θ"}/{(1 + 1/{\text"cos θ"} + {\text"sin θ"}/{\text"cos θ"}) (\text"1 - sin θ")}$

= ${\text"cos θ + 1 - sin θ"}/{{(\text"cos θ + 1 + sin θ") (\text"1 - sin θ")}/{\text"cos θ"}}$

= ${\text"cos θ + 1 - sin θ"}/{{\text"cos θ + 1 + sin θ - sin θ cos θ - sin θ" - sin^2 θ}/{\text"cos θ"}}$

= ${\text"cos θ + 1 - sin θ"}/{{\text"cos θ + 1" - sin^2 θ \text"- sin θ cos θ"}/{\text"cos θ"}}$

(∵ $1 - sin^2 θ = cos^2 θ$)

= ${\text"cos θ + 1 - sin θ"}/{{\text"cos θ +" cos^2 θ \text"- sin θ cos θ"}/{\text"cos θ"}}$

= ${\text"cos θ + 1 - sin θ"}/{{\text"cos (cos θ + 1 - sin θ")}/{\text"cos θ"}}$ = 1