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

Answer: (b)

sin θ = ${2a + 3b}/{3b}$

⇒ sin θ = ${2a}/{3b}$ + 1

as a and b is positive so $(1 + {2a}/{3b})$ will be always greater than 1 that is not possible for sin θ

Option (b) is correct.

Answer: (c)

Statement (1) $√{{\text"1 - cos θ"}/{\text"1 + cos θ"}}$ = cosec θ - cot θ

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

⇒ $√{tan^2 θ/2} = {\text"1 - cos θ"}/{\text"sin θ"} = {2 sin^2 {{θ}/2}}/{2 sin {{θ}/2} cos {{θ}/2}}$

⇒ $tan {θ}/2 = tan {θ}/2$

Hence, (1) is identity

(2) Statement $√{{\text"1 + cos θ"}/{\text"1 - cos θ"}}$ = cosec θ + cot θ

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

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

$√{cot^2 {θ}/2} = cot^2 {θ}/2$

⇒ $cot {θ}/2 = cot {θ}/2$

Hence, (2) is also an identity

∴ Option (c) is correct.

Answer: (b)

Given, ${\text"cos x"}/{\text"cos y"} = n, {\text"sin x"}/{\text"sin y"}$ = m .......(i)

Now, $(m^2 - n^2) sin^2 y = ({sin^2 x}/{sin^2 y} - {cos^2 x}/{cos^2 y})sin^2 y$

= ${(1 - cos^2 x) cos^2 y - cos^2 x (1 - cos^2 y)}/{cos^2 y}$

= ${cos^2 y - cos^2 x}/{cos^2 y} = 1 - n^2$ [from equation (i)]

Answer: (b)

100 θ = 90°

θ = ${90}/{100} = {π}/2 × 1/{100}$

α = cot $(1/{100} . {π}/2) cot (2/{100} . {π}/2)....cot({99}/{100} . {π}/2)$ = 1

Answer: (a)

sin x + $sin^2 x = 1$

sin x = $cos^2x$

∵ 1 - $sin^2 x = cos^2 x$

now

$cos^8x + 2 cos^6 x + cos^4 x$

⇒ $sin^4x + 2 sin^3x + sin^2 x$

⇒ $(sin^2 \text"x + sin x")^2 = 1$