Trigonometric Identities – Objective Type Questions

Q 1 :

If $s e c θ - \tan θ = m$ , then the value of $s e c θ + \tan θ$ is

$1 - \frac{1}{m}$
$m^{2} - 1$
$\frac{1}{m}$
$- m$

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(3)

Given $s e c θ - \tan θ = m$

We have to find the value of $s e c θ + \tan θ .$

We know that

${s e c}^{2} θ - {t a n}^{2} θ = 1 \Rightarrow (s e c θ + t a n θ) (s e c θ - t a n θ) = 1 \Rightarrow (s e c θ + t a n θ) m = 1 [Given] \Rightarrow s e c θ + t a n θ = \frac{1}{m}$

Q 2 :

$I f \frac{x}{3} = 2 s i n A, \frac{y}{3} = 2 c o s A, t h e n t h e v a l u e o f x^{2} + y^{2} i s$

36
9
6
18

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(1)

$G i v e n, \frac{x}{3} = 2 s i n A \Rightarrow s i n A = \frac{x}{6} a n d \frac{y}{3} = 2 c o s A \Rightarrow c o s A = \frac{y}{6} W e k n o w t h a t {s i n}^{2} A + {c o s}^{2} A = 1 \Rightarrow \frac{x^{2}}{36} + \frac{y^{2}}{36} = 1 \Rightarrow x^{2} + y^{2} = 36$

Q 3 :

If $p = c s c θ + c o t θ a n d q = \frac{1}{c s c θ - c o t θ},$ then which of the following is/are correct?

$(i) p - q = 1 (i i) p + q = 1 (i i i) p - q = 0 (i v) p + q = 0$

Choose the correct option from the following:

(i) and (ii)
(ii) and (iii)
(iii) and (iv)
(iii) only

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(4)

We know,

${c s c}^{2} θ - {c o t}^{2} θ = 1 \Rightarrow (c s c θ - c o t θ) (c s c θ + c o t θ) = 1 \Rightarrow c s c θ + c o t θ = \frac{1}{c s c θ - c o t θ} \Rightarrow p = q \Rightarrow p - q = 0$

Q 4 :

If a and b are non-zero positive real numbers and ${c o s}^{2} θ = \frac{(a + b)^{2}}{4 a b}$ $(f o r a l l v a l u e s o f θ),$ then which of the following sets of values are true for a and b?

(i) a = 1, b = 1 and a = 2, b = 2
(ii) a = 3, b = 3 and a = 6, b = 6
(iii) a = 5, b = 5 and a = 7, b = 7
(iv) a = 4, b = 4 and a = 8, b = 8

Choose the correct option from the following:

only (i)
(iii) and (iv)
(ii) and (iii)
All of them

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(4)

We have,

${c o s}^{2} θ = \frac{(a + b)^{2}}{4 a b} \Rightarrow 1 - {c o s}^{2} θ = 1 - \frac{(a + b)^{2}}{4 a b} {[T a k i n g n e g a t i v e a n d t h e n a d d i n g 1 b o t h t h e s i d e s] \Rightarrow {s i n}^{2} θ = \frac{4 a b - (a^{2} + b^{2} + 2 a b)}{4 a b} \Rightarrow {s i n}^{2} θ = \frac{4 a b - a^{2} - b^{2} - 2 a b}{4 a b} \Rightarrow {s i n}^{2} θ = \frac{- a^{2} - b^{2} + 2 a b}{4 a b} = - \frac{(a - b)^{2}}{4 a b}$

Since a, b are positive, the only non-negative value of $- \frac{(a - b)^{2}}{4 a b}$

is zero, which happens when a = b.

Topic Question Set

Q 1 :

If $s e c θ - \tan θ = m$ , then the value of $s e c θ + \tan θ$ is

Q 2 :

$I f \frac{x}{3} = 2 s i n A, \frac{y}{3} = 2 c o s A, t h e n t h e v a l u e o f x^{2} + y^{2} i s$

Q 3 :

If $p = c s c θ + c o t θ a n d q = \frac{1}{c s c θ - c o t θ},$ then which of the following is/are correct?

$(i) p - q = 1 (i i) p + q = 1 (i i i) p - q = 0 (i v) p + q = 0$

Choose the correct option from the following:

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Topic Question Set

Q 1 : If sec θ – tan θ = m, then the value of sec θ + tan θ is .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 2 : If x3=2 sinA, y3=2cosA,then the value of x2+y2 is .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 1 :

If $s e c θ - \tan θ = m$ , then the value of $s e c θ + \tan θ$ is

Q 2 :

$I f \frac{x}{3} = 2 s i n A, \frac{y}{3} = 2 c o s A, t h e n t h e v a l u e o f x^{2} + y^{2} i s$