Test: Inverse Trigonometry- 1 - JEE MCQ

Correct Answer :- b

Explanation : ƒ(x) = √(2tan(x))

x = √(2tan(x))

x² = (2tan(x))

(x²)/2 = tan(x)

tan^-1(x^2)/2 = x

By putting x = √2, we get

x = 1

If 2 tan^-1 (cos x) = tan ^-1(2 cosec x),

2tan^-1(cos x) = tan^-1 (2 cosec x)

= tan^-1(2 cosec x) 

= cot x cosec x = cosec x = x = π/4

tan^-1(-2) + tan^-1(-3)

The value of tan15⁰ + cot 15⁰ 

Hence , the given equation has only one solution.

cot^-1a - cot^-1b + cot^-1 b - cot^-1 c - cot^-1 a = 0

Let y= sin x + cos x

dy/dx=cos x- sin x

For maximum or minimum dy/dx=0

Setting cosx- sin x=0

We get cos x = sin x

x= π/4, 5π/4———-

Whether these correspond to maximum or minimum, can be found from the sign of second derivative.

d^2y/dx^2=-sin x - cos x=-1/√2–1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5π/4

d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5π/4 corresponds to minimum

Maximum value of the function

y= sin π/4 + cos π/4= 2/√2=√2

Minimum value is

Sin(5π/4)+cos (5π/4)=-2/√2=-√2

Because both sin 200⁰ and cos 200⁰ lies in 3rd quadrant. In 3rd quadrant values of sin and cos are negative.

If cos^(-1)x + cos^(-1) y = 2π, then the value of sin^(-1)x + sin^(-1)y  = π−2π = −π.

Since sin⁻¹x is defined for |x|⩽1, and sec⁻¹x is defined for |x|⩾1,therefore,f(x) is defined only when|x|=1.so, D_f = {−1,1}.

Put   therefore the given expressionis sin2θ = 2sinθcosθ

= [(5/4) + (3/4)] / [(5/4) - (3/4)]
=(8/4) / (2/4)
=4

(sinA+cosA)²
= sin²A+cos²A+2sinAcosA
=11+sin²A=1sin²A=0.
(because Sin 2A = 2sin A cos A)

Therefore, tanθ =

As no value of x in (0, 1) can satisfy the given equation.Thus, the given equation has only one solution.

Put

sin^-1 x = θ ⇒ x = sin θ

Put,
cos^-1x = θ ⇒ x = cos θ ⇒ cos θ