trigonometric ratios and identity Model Questions & Answers, Practice Test for ssc mts paper 1 2023

Answer: (c)

$9 tan^2 θ + 4 cot^2 θ$

⇒ $(\text"3 tan θ")^2 + (\text"2 cot θ")^2 - 2 (\text"3 tan θ") \text"(2) cot θ + 2"(\text"3 tan θ")^2 cot θ$

⇒ $(\text"3 tan θ - 2 cot θ")^2 + 12$

since $(\text"3 tan θ - 2 cot θ")^2 ≥ 0$

Minimum value is 12

Answer: (d)

I. $tan^2 θ - sin^2 θ = {sin^2 θ}/{cos^2 θ} - sin^2 θ, θ ≠ (2n + 1) {π}/2$

= ${sin^2 θ (1 - cos^2 θ)}/{cos^2 θ}, θ ≠ (2n + 1) {π}/2$

= ${sin^2 θ}/{cos^2 θ} sin^2 θ, θ ≠ (2n + 1) {π}/2$

= $tan^2 θ sin^2 θ, θ ≠ (2n + 1) {π}/2$

II. (cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ)

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

[θ ≠ n π (2n + 1)${π}/2$]

= ${cos^2 θ sin^2 θ sec^2 θ}/{sin θ cos θ tan θ}, θ ≠ n π, (2n + 1) {π}/2$

= sin θ cos θ $1/{cos^2 θ} . {\text"cos θ"}/{\text"sin θ"}, θ ≠ n π, (2n + 1) {π}/2$

= 1

Since, to become an identity, both statements must be satisfied for every value of &theta.

Therefore, neither I not II are the identities.

Answer: (a)

Let angles in circular measures are A and B, then

degree measures will be ${π A}/{180°} and {π B}/{180°}$

According to question,

A + B = 1 .......(i)

and ${π A}/{180°} - {π B}/{180°}$ = 1 ......(ii)

On solving equations (i) and (ii), we get

A = ${90}/{π} ({π}/{180} + 1) ⇒ A = (1/2 + {90}/{π})$

From equation (i),

$1/2 + {90}/{π} + B = 1 ⇒ B - (1/2 - {90}/{π})$

Answer: (b)

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

= ${2 sin {θ}/2 . cos {θ}/2}/{1 + 2 cos^2 {θ}/2 - 1} + {1 + 2 cos^2 {θ}/2 - 1}/{2 sin {θ}/2 . cos {θ}/2}$

= ${sin {θ}/2}/{cos {θ}/2} + {cos {θ}/2}/{sin {θ}/2}$

= $2/2 . {sin^2 {θ}/2 + cos^2 {θ}/2}/{sin {θ}/2 . cos {θ}/2} = 2/{sin θ}$ = 2 cosec θ

Answer: (c)

Given, $\text"tan θ" = p/q$

⇒ sec θ = ${√{p^2 + q^2}}/q \text"and cosec θ" = {√{p^2 + q^2}}/p$

∴ ${\text"p sec θ - q cosec θ"}/{\text"p sec θ + q cosec θ"}$

= ${p({√{p^2 + q^2}}/q) - q({√{p^2 + q^2}}/p)}/{p({√{p^2 + q^2}}/q) + q ({√{p^2 + q^2}}/p)}$

= ${p/q - q/p}/{p/q + q/p} = {p^2 - q^2}/{p^2 + q^2}$