JEE Advanced Maths PYP Inverse Trigonometric Functions

Q 1 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\tan (\sin^{- 1} (\frac{3}{5}) - 2 \cos^{- 1} (\frac{2}{\sqrt{5}}))$ is [2024]

$\frac{7}{24}$
$\frac{- 7}{24}$
$\frac{- 5}{24}$
$\frac{5}{24}$

1

2

3

4

(2)

$We have, \tan (\sin^{- 1} (\frac{3}{5}) - 2 \cos^{- 1} (\frac{2}{\sqrt{5}}))$

$Let \sin^{- 1} (\frac{3}{5}) = α, 2 \cos^{- 1} (\frac{2}{\sqrt{5}}) = β \Rightarrow \cos \frac{β}{2} = \frac{2}{\sqrt{5}}$

$∵$ $\sin α = \frac{3}{5} \Rightarrow \tan α = \frac{3}{4}$

$\tan β = \frac{2 \tan \frac{β}{2}}{1 - \tan^{2} \frac{β}{2}} = \frac{2 \times \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}$

$\Rightarrow \tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β} = \frac{\frac{3}{4} - \frac{4}{3}}{1 + 1} = - \frac{7}{24}$

Q 2 :

The value of $\sec^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \sec (\frac{7 π}{12} + \frac{k π}{2}) \sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$ in the interval $[- \frac{π}{4}, \frac{3 π}{4}]$ equals _______ . [2019]

(0)

$\sec^{- 1} [\frac{1}{4} \sum_{k = 0}^{10} \sec (\frac{7 π}{12} + \frac{k π}{2}) \sec (\frac{7 π}{12} + \frac{(k + 1) π}{2})]$

$= \sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{1}{2 \cos (\frac{7 π}{12} + \frac{k π}{2}) \cos (\frac{7 π}{12} + \frac{k π}{2} + \frac{π}{2})}]$

$= \sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin (\frac{7 π}{6} + k π)}]$

$= \sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin ((k + 1) π + \frac{π}{6})}]$

If $k$ is an even integer, then

$\sin ((k + 1) π + \frac{π}{6}) = - \sin \frac{π}{6} = \frac{- 1}{2}$

If $k$ is an odd integer, then $\sin ((k + 1) π + \frac{π}{6}) = \sin \frac{π}{6} = \frac{1}{2}$

$∴$ $\sum_{k = 0}^{9} \frac{1}{\sin ((k + 1) π + \frac{π}{6})} = 0$

Hence, $\sec^{- 1} [\frac{1}{2} \sum_{k = 0}^{10} \frac{- 1}{\sin ((k + 1) π + \frac{π}{6})}]$

$= \sec^{- 1} [\frac{1}{2} (\frac{- 1}{- \frac{1}{2}})] = \sec^{- 1} (1) = 0$

Q 3 :

If $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

$\cos β > 0$
$\sin β < 0$
$\cos (α + β) > 0$
$\cos α < 0$

1

2

3

4

Select one or more options

(2, 3, 4)

$α = 3 \sin^{- 1} (\frac{6}{11}) > 3 \sin^{- 1} (\frac{1}{2}) = \frac{π}{2} \Rightarrow α > \frac{π}{2}$

$∴$ $\cos α < 0$

$β = 3 \cos^{- 1} (\frac{4}{9}) > 3 \cos^{- 1} (\frac{1}{2}) = π \Rightarrow β > π$

$∴$ $\cos β < 0$ and $\sin β < 0$

Now, $α + β > \frac{3 π}{2},$ $∴$ $\cos (α + β) > 0$

Q 4 :

Match List I with List II and select the correct answer using the code given below the lists : [2013]

List I List II

P. $(\frac{1}{y^{2}} {{(\frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)})}^{2} + y^{4})}^{1 / 2}$ takes value 1. $\frac{1}{2} \sqrt{\frac{5}{3}}$

Q. If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z,$
then possible value of $\cos (\frac{x - y}{2})$ is 2. $\sqrt{2}$

R. If $\cos (\frac{π}{4} - x) \cos 2 x + \sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x + \cos (\frac{π}{4} + x) \cos 2 x,$
then possible value of $\sec x$ is 3. $\frac{1}{2}$

S. If $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} (x \sqrt{6})),$ $x \neq 0,$ then possible value of $x$ is 4. $1$

Codes:

P - 4, Q - 3, R - 1, S - 2
P - 4, Q - 3, R - 2, S - 1
P - 3, Q - 4, R - 2, S - 1
P - 3, Q - 4, R - 1, S - 2

1

2

3

4

(2)

$(P) {[\frac{1}{y^{2}} {(\frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)})}^{2} + y^{4}]}^{\frac{1}{2}}$

$= {[\frac{1}{y^{2}} {(\frac{\cos (\cos^{- 1} \frac{1}{\sqrt{1 + y^{2}}}) + y \sin (\sin^{- 1} \frac{y}{\sqrt{1 + y^{2}}})}{\cot (\cot^{- 1} \frac{\sqrt{1 - y^{2}}}{y}) + \tan (\tan^{- 1} \frac{y}{\sqrt{1 - y^{2}}})})}^{2} + y^{4}]}^{\frac{1}{2}}$

$= {[\frac{1}{y^{2}} {(\frac{\sqrt{1 + y^{2}}}{\frac{1}{\frac{1}{y \sqrt{1 - y^{2}}}}})}^{2} + y^{4}]}^{\frac{1}{2}}$

$= {(1 - y^{4} + y^{4})}^{\frac{1}{2}} = 1$ $∴$ $(P) \to (4)$

$(Q) \cos x + \cos y = - \cos z$ $\dots (i)$

$and \sin x + \sin y = - \sin z$ $\dots (i i)$

On squaring (i) and (ii) and then adding, we get

${(\cos x + \cos y)}^{2} + {(\sin x + \sin y)}^{2} = \cos^{2} z + \sin^{2} z$

$\Rightarrow 2 + 2 \cos (x - y) = 1$

$\Rightarrow 4 \cos^{2} \frac{x - y}{2} = 1$ $\Rightarrow \cos \frac{x - y}{2} = \pm \frac{1}{2}$

$∴$ $(Q) \to (3)$

$(R) \cos (\frac{π}{4} - x) \cos 2 x + \sin x \sin 2 x \sec x$

$= \cos x \sin 2 x \sec x + \cos (\frac{π}{4} + x) \cos 2 x$

$\Rightarrow \cos 2 x [\cos (\frac{π}{4} - x) - \cos (\frac{π}{4} + x)] = \sin 2 x \sec x (\cos x - \sin x)$

$\Rightarrow 2 \sin \frac{π}{4} \sin x \cos 2 x = 2 \sin x (\cos x - \sin x)$

$\Rightarrow 2 \sin x [\frac{1}{\sqrt{2}} (\cos^{2} x - \sin^{2} x) - (\cos x - \sin x)] = 0$

$\Rightarrow 2 \sin x (\cos x - \sin x) (\frac{\cos x + \sin x}{\sqrt{2}} - 1) = 0$

$\Rightarrow \sin x = 0 or \tan x = 1 or \cos (x - \frac{π}{4}) = 1$

$\Rightarrow x = 0 or \frac{π}{4} \Rightarrow \sec x = 1 or \sqrt{2}$

$∴$ $(R) \to (2, 4)$

$(S) \cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} x \sqrt{6})$

$\Rightarrow \frac{x}{\sqrt{1 - x^{2}}} = \frac{x \sqrt{6}}{\sqrt{1 + 6 x^{2}}}$ $\Rightarrow x = \pm \frac{1}{2} \sqrt{\frac{5}{3}}$

$∴$ $(S) \to (1)$

Hence, $(P) \to (4), (Q) \to (3), (R) \to (2, 4), (S) \to (1)$

Topic Question Set

Q 1 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\tan (\sin^{- 1} (\frac{3}{5}) - 2 \cos^{- 1} (\frac{2}{\sqrt{5}}))$ is [2024]

Q 2 :

The value of $\sec^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \sec (\frac{7 π}{12} + \frac{k π}{2}) \sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$ in the interval $[- \frac{π}{4}, \frac{3 π}{4}]$ equals _______ . [2019]

Q 3 :

If $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

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	List I		List II
P.	$(\frac{1}{y^{2}} {{(\frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)})}^{2} + y^{4})}^{1 / 2}$ takes value	1.	$\frac{1}{2} \sqrt{\frac{5}{3}}$
Q.	If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z,$ then possible value of $\cos (\frac{x - y}{2})$ is	2.	$\sqrt{2}$
R.	If $\cos (\frac{π}{4} - x) \cos 2 x + \sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x + \cos (\frac{π}{4} + x) \cos 2 x,$ then possible value of $\sec x$ is	3.	$\frac{1}{2}$
S.	If $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} (x \sqrt{6})),$ $x \neq 0,$ then possible value of $x$ is	4.	$1$

Topic Question Set

Q 1 : Considering only the principal values of the inverse trigonometric functions, the value of tan(sin-1(35)-2cos-1(25)) is [2024]

Q 2 : The value of sec-1(14∑k=010sec(7π12+kπ2)sec(7π12+(k+1)π2)) in the interval [-π4,3π4] equals _______ . [2019]

Q 3 : If α=3sin-1(611) and β=3cos-1(49), where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

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Q 1 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\tan (\sin^{- 1} (\frac{3}{5}) - 2 \cos^{- 1} (\frac{2}{\sqrt{5}}))$ is [2024]

Q 2 :

The value of $\sec^{- 1} (\frac{1}{4} \sum_{k = 0}^{10} \sec (\frac{7 π}{12} + \frac{k π}{2}) \sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$ in the interval $[- \frac{π}{4}, \frac{3 π}{4}]$ equals _______ . [2019]

Q 3 :

If $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]