trigonometric ratios and identity Model Questions & Answers, Practice Test for ssc mts paper 1 2023
ssc mts paper 1 2023 SYLLABUS WISE SUBJECTS MCQs
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Trigonometric Ratios & Identity
If a and b are positive, then the relation sinθ = ${2a + 3b}/{3b}$ is
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.
Consider the following :
1. $√{{1 - cos θ}/{1 + cos θ}}$ = cosec θ - cot θ
2. $√{{1 + cos θ}/{1 - cos θ}}$ = cosec θ + cot θ
Which of the above is/are identity identities?
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.
If ${\text"cos x"}/{\text"cos y"}$ = n and ${\text"sin x"}/{\text"sin y"}$ = m, then $(m^2 - n^2) sin^2$ y is equal to
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)]
Let 0 < θ < 90° and 100 θ = 90°. If α = $Π^{99}_{n = 1}$ cot nθ, then which one of the following is correct ?
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
If sinx + $sin^2$ x = 1, then what is the value of $cos^8x + 2cos^6x + cos^4$x ?
Answer: (a)
sin x + $sin^2 x = 1$
sin x = $cos^2x$
∵ 1 - $sin^2 x = cos^2 x$
now
$cos^8x + 2 cos^6 x + cos^4 x$
⇒ $sin^4x + 2 sin^3x + sin^2 x$
⇒ $(sin^2 \text"x + sin x")^2 = 1$
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