trigonometric ratios and identity Model Questions & Answers, Practice Test for ibps clerk prelims 2023
ibps clerk prelims 2023 SYLLABUS WISE SUBJECTS MCQs
Ratio & Proportion
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Trigonometric Ratios & Identity
Linear Equations
Quadratic Equations
Logarithm
If p = a sin x + b cos x and q = a cos x – b sin x, then what is the value of $p^2 + q^2$ ?
Answer: (c)
Here, p = a sin x + b cos x and q = a cos x - b sin x
On squaring both sides,
⇒ $p^2 = a^2 sin^2 x + b^2 cos^2 x$ + 2ab sin x cos x ......(i)
q = a cos x - b sin x
On squaring both sides,
and $q^2 = a^2 cos^2 x + b^2 sin^2$ x - 2ab sin x cos x .......(ii)
Now, add equation (i) and equation (ii), we get
∴ $p^2 + q^2 = a^2 (sin^2 x + cos^2 x) + b^2 (cos^2 x + sin^2 x)$
= $a^2 + b^2$
If x + $(1/x)$ = 2 cos α, then what is the value of $x^2 + (1/{x^2})$ ?
Answer: (c)
x + $1/x$ = 2 cos α
Squaring both sides, then we get
$x^2 + 1/{x^2} + 2 = 4 cos^2 α$
⇒ $x^2 + 1/{x^2} = 2(2 cos^2 α -1)$
=2 $(2 cos^2 α - sin^2 α - cos^2 α)$
= $2 cos^2 α - 2 sin^2 α$
Assertion (A) : tan 50° > 1.
Reason (R) : tan θ > 1 for 0° < θ < 90°.
Answer: (c)
We know that, tan θ is increasing in 0° to 90° and tan 45° = 1.
∴ 50° > 1.
So, A is true but R is false.
If θ lies in the first quadrant and cot θ = ${63}/{16}$, then what is the value of (sin θ + cos θ) ?
Answer: (c)
As cot θ = ${63}/{16}$
From Δ ABC
B = 63, P = 16
∴ H = $√{63^2 + 16^2}$ = 65
Now, sin θ + cos θ
= ${P + B}/H = {63 + 16}/{65} = {79}/{65}$
If α and β are complimentary angles, then what is $√{\text"cosec α . cosec β"}({\text"sin α"}/{\text"sin β"} + {\text"cos α"}/{\text"cos β"})^{- 1/2}$ equal to?
Answer: (a)
Given, α + β = 90°......(i)
∴ $√{\text"cosec α . cosec β"} ({\text"sin α"}/{\text"sin β"} + {\text"cos α"}/{\text"cos β"})^{- 1/2}$
= $1/{(\text"sin α sin β")^{1/2}} ({\text"sin α cos β + cos α sin β"}/{\text"sin β cos β"})^{- 1/2}$
= $1/{(\text"sin α sin β")^{1/2}} ({sin (α + β)}/{\text"sin β cos β"})^{- 1/2}$
= $1/{(\text"sin α sin β")^{1/2}} ({\text"sin 90°"}/{\text"cos (90° - α) sin β"})^{- 1/2}$
[from equation (i)]
= $1/{(\text"sin α sin β")^{1/2}} × (\text"sin α sin β")^{1/2}$ = 1
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