Free Inverse Trigonometric Functions 03 Practice Test - 12th Grade - Commerce

Question 1

If $x = s i n (2 t a n^{- 1} 2), y = s i n (\frac{1}{2} t a n^{- 1} \frac{4}{3})$ then

A. x = 1 – y

x^{2} = 1 - y

x^{2} = 1 + y

y^{2} = 1 - x

SOLUTION

Solution : D

Let $t a n^{- 1} 2 = α \Rightarrow x = s i n 2 α = \frac{4}{5} y = s i n (\frac{β}{2}) = \sqrt{\frac{1 - c o s β}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}$

Question 2

$s i n^{- 1} | c o s x | - c o s^{- 1} | s i n x | = a$ has at least one solution if a $\in$

0

[0, \frac{π}{2}]

[\frac{π}{2}, \frac{3 π}{2}]

(0, π)

SOLUTION

Solution : A

$s i n^{- 1} | c o s x | - c o s^{- 1} | s i n x | = \frac{π}{2} - c o s^{- 1} | c o s x | - \frac{π}{2} + s i n^{- 1} | s i n x | = a \Rightarrow s i n^{- 1} | s i n x | - c o s^{- 1} | c o s x | = a \Rightarrow a = 0 \forall x$

Question 3

${sin}^{- 1} (sin 10) =$

A. 10

3 π - 10

π + 6

2 π - 6

SOLUTION

Solution : B

$3 π < 10 < 3 π + \frac{π}{2} ∴ 10 rad \in Q_{3} ∴ - \frac{π}{2} < 3 π - 10 < 0 ∴ 3 π - 10 \in Q_{4} Also sin (3 π - 10) = sin 10 Hence {sin}^{- 1} (sin 10) = {sin}^{- 1} sin (3 π - 10) = 3 π - 10$

Question 4

$c o s^{- 1} x = t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$ , then:

X \in R

- \leq x \leq 1, x \neq 0

0 < x \leq 1

N o n e o f t h e s e

SOLUTION

Solution : C

Putting $θ = c o s^{- 1} x$ in R.H.S., we have
R.H.S $= t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x} = t a n^{- 1} \frac{\sqrt{1 - c o s^{2} θ}}{c o s θ} (0 \leq θ \leq π) = t a n^{- 1} \frac{s i n θ}{c o s θ} = t a n^{- 1} t a n θ = θ = c o s^{- 1} x$ when $- \frac{π}{2} < θ < \frac{π}{2}$
i.e., when $0 \leq θ < \frac{π}{2}$
i.e., when $0 < c o s θ \leq 1$ i.e., $0 < x \leq 1$ .

Question 5

The value of x for which $s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x)$ is:

\frac{1}{2}

B. 1

C. 0

- \frac{1}{2}

SOLUTION

Solution : D

$s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x) \Rightarrow c o t^{- 1} (1 + x) = [\frac{π}{2} \pm t a n^{- 1} x] \Rightarrow \frac{π}{2} - t a n^{- 1} (1 + x) = \frac{π}{2} \pm t a n^{- 1} x \Rightarrow 1 + x = \pm x$
$x = - \frac{1}{2}$ Which, on verification, satisfies the equation.

Question 6

If $θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}$ , where $a_{1}, a_{2}, a_{3}, \dots a_{n}$ are in A.P. with common difference d, then $t a n θ =$

\frac{(n - 1) d}{1 + a_{1} a_{n}}

\frac{(n - 1) d}{a_{1} + a_{n}}

\frac{a_{n} - a_{1}}{a_{n} + a_{1}}

\frac{n d}{1 + a_{1} a_{n}}

SOLUTION

Solution : A

$θ = t a n^{- 1} \frac{a_{2} - a_{1}}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{a_{3} - a_{2}}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}}$
$= (t a n^{- 1} a_{2} - t a n^{- 1} a_{1}) + (t a n^{- 1} a_{3} - t a n^{- 1} a_{2}) + \dots + t a n^{- 1} \frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}} + \dots + (t a n^{- 1} a_{n} - t a n^{- 1} a_{n - 1})$
$= t a n^{- 1} a_{n} - t a n^{- 1} a_{1}$
$= t a n^{- 1} \frac{a_{n} - a_{1}}{1 + a_{1} a_{n}} = t a n^{- 1} \frac{(n - 1) d}{1 + a_{1} a_{n}}$
$∴ t a n θ = \frac{(n - 1) d}{1 + a_{1} a_{n}}$

Question 7

$c o s [c o s^{- 1} (\frac{- 1}{7}) + s i n^{- 1} (\frac{- 1}{7})] =$

\frac{- 1}{3}

0

\frac{1}{3}

\frac{4}{9}

SOLUTION

Solution : B

$c o s {c o s^{- 1} (\frac{- 1}{7}) + s i n^{- 1} (\frac{- 1}{7})} = c o s \frac{π}{2} = 0$

Question 8

If $c o s^{- 1} \sqrt{p} + c o s^{- 1} \sqrt{1 - p} + c o s^{- 1} \sqrt{1 - q} = \frac{3 π}{4}$ , then the value of q is

1

\frac{1}{\sqrt{2}}

\frac{1}{3}

\frac{1}{2}

SOLUTION

Solution : D

Let $α = c o s^{- 1} \sqrt{p} : β = c o s^{- 1} \sqrt{1 - p}$
and $γ = c o s^{- 1} \sqrt{1 - q}$
or
$c o s α = \sqrt{p} : c o s β = \sqrt{1 - p}$
and $c o s γ = \sqrt{1 - q} .$
Therefore $s i n α = \sqrt{1 - p}, s i n β = \sqrt{p}$ and $s i n γ = \sqrt{q} .$
The given equation may be written as $α + β + γ = \frac{3 π}{4}$
$o r, α + β = \frac{3 π}{4} - γ$
or, $c o s (α + β) = c o s (\frac{3 π}{4} - γ)$
$\Rightarrow c o s α c o s β - s i n α s i n β$ = $c o s {π - (\frac{π}{4} + γ)} = - c o s (\frac{π}{4} + γ)$
$\Rightarrow \sqrt{p} \sqrt{1 - p} - \sqrt{1 - p} \sqrt{p} = - (\frac{1}{\sqrt{2}} \sqrt{1 - q} - \frac{1}{\sqrt{2}} . \sqrt{q})$
$\Rightarrow 0 = \sqrt{1 - q} - \sqrt{q} \Rightarrow 1 - q = q \Rightarrow q = \frac{1}{2}$

Question 9

$c o t^{- 1} [(c o s α)^{\frac{1}{2}}] - t a n^{- 1} [(c o s α)^{\frac{1}{2}}] = x$ , then $s i n x =$

t a n^{2} (\frac{α}{2})

c o t^{2} (\frac{α}{2})

t a n α

c o t (\frac{α}{2})

SOLUTION

Solution : A

$t a n^{- 1} [\frac{1}{\sqrt{c o s α}}] - t a n^{- 1} [\sqrt{c o s α}] = x$
$\Rightarrow t a n^{- 1} ⎡ ⎢ ⎣ \frac{\frac{1}{\sqrt{c o s α}} - \sqrt{c o s α}}{1 + \frac{\sqrt{c o s α}}{\sqrt{c o s α}}} ⎤ ⎥ ⎦ = x \Rightarrow t a n x = \frac{1 - c o s α}{2 \sqrt{c o s α}}$
$∴ s i n x = \frac{1 - c o s α}{1 + c o s α} = \frac{2 s i n^{2} \frac{α}{2}}{2 c o s^{2} \frac{α}{2}} = t a n^{2} (\frac{α}{2})$ .

Question 10

If $t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$ , then $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} =$

0

1

\frac{1}{x y z}

x y z

SOLUTION

Solution : B

$t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$
$\Rightarrow t a n^{- 1} x + t a n^{- 1} y = π - t a n^{- 1} z$
$\Rightarrow \frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z + x y z$
$\Rightarrow x + y + z = x y z$
Dividing by xyz, we get
$\frac{1}{y z} + \frac{1}{x z} + \frac{1}{x y} = 1.$
Note: Students should remember this question as a formula.

Free Inverse Trigonometric Functions 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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