JEE Main Previous Year Trigonometry Questions and Solutions

Q 11 :

The value of $\sin 12 ° \sin 48 ° \sin 54 °$ is equal to

$\frac{1}{16}$
$\frac{1}{32}$
$\frac{1}{8}$
$\frac{1}{4}$

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(3)

Now, $\sin 12 ° \sin 48 ° \sin 54 ° = \frac{1}{2} (\cos 36 ° - \cos 60 °) \cos 36 °$

$= \frac{1}{2} [\frac{\sqrt{5} + 1}{4} - \frac{1}{2}] [\frac{\sqrt{5} + 1}{4}]$

$= \frac{1}{2} [\frac{\sqrt{5} - 1}{4}] [\frac{\sqrt{5} + 1}{4}]$

$= \frac{5 - 1}{32} = \frac{4}{32} = \frac{1}{8}$

Q 12 :

The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is

cot A
tan A
cos A
sin A

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(1)

$\tan A + 2 \tan 2 A + 4 \tan 4 A + 8 (\frac{1 - \tan^{2} 4 A}{2 \tan 4 A})$

$= \tan A + 2 \tan 2 A + (\frac{4 \tan^{2} 4 A + 4 - 4 \tan^{2} 4 A}{\tan 4 A})$

$= \tan A + 2 \tan 2 A + 4 \cot 4 A$

$= \tan A + 2 \tan 2 A + 4 (\frac{1 - \tan^{2} 2 A}{2 \tan 2 A})$

$= \tan A + [\frac{2 \tan^{2} 2 A + 2 - 2 \tan^{2} 2 A}{\tan 2 A}]$

$= \tan A + 2 \cot 2 A$

$= \tan A + 2 (\frac{1 - \tan^{2} A}{2 \tan A})$

$= \frac{\tan^{2} A + 1 - \tan^{2} A}{\tan A} = \cot A$

Q 13 :

The maximum value of $\cos^{2} (\frac{π}{3} - x) - \cos^{2} (\frac{π}{3} + x)$ is

$- \frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{3}{2}$

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(3)

$\cos^{2} (\frac{π}{3} - x) - \cos^{2} (\frac{π}{3} + x)$

$= [\cos (\frac{π}{3} - x) + \cos (\frac{π}{3} + x)] [\cos (\frac{π}{3} - x) - \cos (\frac{π}{3} + x)]$

$= (2 \cos \frac{π}{3} \cos x) (2 \sin \frac{π}{3} \sin x)$

$= \sin \frac{2 π}{3} \sin 2 x = \frac{\sqrt{3}}{2} \sin 2 x$

Hence, maximum value of the given expression is $\frac{\sqrt{3}}{2} .$

Q 14 :

If sum of all the solution of the equation $8 \cos x (\cos (\frac{π}{6} + x) \cdot \cos (\frac{π}{6} - x) - \frac{1}{2}) = 1$ in $[0, π]$ is $k π$ , then $k$ is equal to

$\frac{2}{3}$
$\frac{13}{9}$
$\frac{8}{9}$
$\frac{20}{9}$

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(2)

Key idea Apply the identities

$\cos (x + y) \cos (x - y) = \cos^{2} x - \sin^{2} y$ and $\cos 3 x = 4 \cos^{3} x - 3 \cos x$

We have,

$8 \cos x (\cos (\frac{π}{6} + x) \cos (\frac{π}{6} - x) - \frac{1}{2}) = 1$

$\Rightarrow 8 \cos x (\cos^{2} \frac{π}{6} - \sin^{2} x - \frac{1}{2}) = 1$

$\Rightarrow$ $8 \cos x (\frac{3}{4} - \sin^{2} x - \frac{1}{2}) = 1$

$\Rightarrow 8 \cos x (\frac{3}{4} - \frac{1}{2} - 1 + \cos^{2} x) = 1$

$\Rightarrow$ $8 \cos x (\frac{- 3 + 4 \cos^{2} x}{4}) = 1$

$\Rightarrow$ $2 (4 \cos^{3} x - 3 \cos x) = 1$

$\Rightarrow$ $2 \cos 3 x = 1 \Rightarrow \cos 3 x = \frac{1}{2}$

$\Rightarrow$ $3 x = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} [0 \leq 3 x \leq 3 π]$

$\Rightarrow$ $x = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}$

Now, $Sum = \frac{π}{9} + \frac{5 π}{9} + \frac{7 π}{9} = \frac{13 π}{9}$

$\Rightarrow$ $k π = \frac{13 π}{9}$

$∴$ $k = \frac{13}{9}$

Q 15 :

If $\sin^{2} θ = \frac{4 x y}{{(x + y)}^{2}}$ is true, if and only if

$x - y \neq 0$
$x = - y$
$x + y \neq 0$
$x \neq 0, y \neq 0$

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(3)

$∵$ $\sin^{2} θ \leq 1$

$∴$ $\frac{4 x y}{{(x + y)}^{2}} \leq 1$

$\Rightarrow$ $x^{2} + y^{2} + 2 x y - 4 x y \geq 0$

$\Rightarrow$ ${(x - y)}^{2} \geq 0$

which is true for all real values of $x$ and $y$ , provided $x + y \neq 0$ , otherwise $\frac{4 x y}{{(x + y)}^{2}}$ will be meaningless.

Q 16 :

If $A = \sin^{2} x + \cos^{4} x,$ then for all real $x$

$\frac{13}{16} \leq A \leq 1$
$1 \leq A \leq 2$
$\frac{3}{4} \leq A \leq \frac{13}{16}$
$\frac{3}{4} \leq A \leq 1$

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(4)

$A = \sin^{2} x + \cos^{4} x$

$\Rightarrow$ $A = 1 - \cos^{2} x + \cos^{4} x$

$=$ $\cos^{4} x - \cos^{2} x + \frac{1}{4} + \frac{3}{4}$

$=$ ${(\cos^{2} x - \frac{1}{2})}^{2} + \frac{3}{4}$ $\dots (i)$

where, $0 \leq {(\cos^{2} x - \frac{1}{2})}^{2} \leq \frac{1}{4}$ $\dots (i i)$

$∴$ $\frac{3}{4} \leq A \leq 1$

Q 17 :

If $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x),$ where $x \in R, k \geq 1$ , then $f_{4} (x) - f_{6} (x)$ is equal to

$\frac{1}{6}$
$\frac{1}{3}$
$\frac{1}{4}$
$\frac{1}{12}$

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(4)

Given, $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x),$ where $x \in R$ and $k \geq 1$

Now, $f_{4} (x) - f_{6} (x)$

$= \frac{1}{4} (\sin^{4} x + \cos^{4} x) - \frac{1}{6} (\sin^{6} x + \cos^{6} x)$

$= \frac{1}{4} (1 - 2 \sin^{2} x \cdot \cos^{2} x) - \frac{1}{6} (1 - 3 \sin^{2} x \cdot \cos^{2} x)$

$= \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$

Q 18 :

For $0 < ϕ < \frac{π}{2}$ , if $x = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ, y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ \sin^{2 n} ϕ,$ then

$x y z = x z + y$
$x y z = x y - z$
$x y z = x + y + z$
$x y z = y z + x$

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(3)

$x = \frac{1}{1 - \cos^{2} ϕ} = \frac{1}{\sin^{2} ϕ}, y = \frac{1}{\cos^{2} ϕ}$

and $z = \frac{1}{1 - \sin^{2} ϕ \cdot \cos^{2} ϕ}$

Clearly, $\frac{1}{x} + \frac{1}{y} = 1$

$\Rightarrow$ $x y = x + y$ and $\frac{1}{z} = 1 - \frac{1}{x y}$

$\Rightarrow$ $x y = x y z - z$

$∴$ $x y z = x y + z = x + y + z$

Q 19 :

The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

infinite number of real roots
no real roots
exactly one real root
exactly four real roots

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(2)

Given equation is

$e^{\sin x} - e^{- \sin x} = 4$ $\Rightarrow e^{\sin x} - \frac{1}{e^{\sin x}} = 4$

Now, let $y = e^{\sin x}$

Then, we get

$y - \frac{1}{y} = 4$ $\Rightarrow y^{2} - 4 y - 1 = 0$

$∴$ $y = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}$

$\Rightarrow e^{\sin x} = 2 \pm \sqrt{5}$

Since sine is a bounded function, i.e., $- 1 \leq \sin x \leq 1 .$

Therefore, we get $e^{- 1} \leq e^{\sin x} \leq e$

$\Rightarrow e^{\sin x} \in [\frac{1}{e}, e]$

Also, it is obvious that $2 + \sqrt{5} > e$

and $2 - \sqrt{5} < \frac{1}{e}$ $\Rightarrow 2 \pm \sqrt{5} \notin [\frac{1}{e}, e]$

So, $e^{\sin x} = 2 + \sqrt{5}$ is not possible for any $x \in R$ ,

and $e^{\sin x} = 2 - \sqrt{5}$ is also not possible for any $x \in R$

Hence, we can say that the given equation has no solution.

Q 20 :

The angles of a triangle are in A.P. The number of degrees in the least is to be the number of radians in the greatest as $60 : π$ . Then the greatest angle is

$120 °$
$90 °$
$135 °$
$105 °$

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(2)

Let the angles of the triangle be $(a - d) °, a °, (a + d) °$

Their sum is $a - d + a + a + d = 180$

$\Rightarrow 3 a = 180$ $\Rightarrow a = 60$

Again, by the given condition,

$\frac{a - d}{\frac{(a + d) π}{180}} = \frac{60}{π}$ $\Rightarrow \frac{3 (a - d)}{a + d} = 1$

$\Rightarrow 3 a - 3 d = a + d$

$\Rightarrow 2 a = 4 d$ $\Rightarrow$ $a = 2 d$

$∴$ $60 = 2 d$ $∴$ $d = 30$

Hence, the greatest angle $= 60 + 30 = 90 °$

Q 21 :

The value of $6 (\sin^{6} θ + \cos^{6} θ) - 9 (\sin^{4} θ + \cos^{4} θ) + 4$ is

(1)

$6 (\sin^{6} θ + \cos^{6} θ) - 9 (\sin^{4} θ + \cos^{4} θ) + 4$

$= 6 [{(\sin^{2} θ + \cos^{2} θ)}^{3} - 3 \sin^{2} θ \cos^{2} θ (\sin^{2} θ + \cos^{2} θ)] - 9 [{(\sin^{2} θ + \cos^{2} θ)}^{2} - 2 \sin^{2} θ \cos^{2} θ] + 4$

$= 6 [1 - 3 \sin^{2} θ \cos^{2} θ] - 9 (1 - 2 \sin^{2} θ \cos^{2} θ) + 4$

$= 6 - 9 + 4 = 1$

Q 22 :

The value of $\tan 15 ° + \cot 15 °$ is

(4)

$\tan 15 ° + \cot 15 ° = \tan 15 ° + \frac{1}{\tan 15 °}$

$= \frac{1 + \tan^{2} 15 °}{\tan 15 °}$

$= 2 [\frac{1}{\frac{2 \tan 15 °}{1 + \tan^{2} 15 °}}]$

$= \frac{2}{\sin 30 °} = \frac{2}{\frac{1}{2}} = 4$

Q 23 :

If $\sin θ_{1} + \sin θ_{2} + \sin θ_{3} = 3,$ then $\cos θ_{1} + \cos θ_{2} + \cos θ_{3} =$

(0)

Since $\sin θ \in [- 1, 1]$

and $\sin θ_{1} + \sin θ_{2} + \sin θ_{3} = 3$

$\Rightarrow$ $each \sin θ = 1$

$∴$ $each \cos θ = 0$

Hence, $\cos θ_{1} + \cos θ_{2} + \cos θ_{3} = 0$

Q 24 :

The value of $\frac{\cot 54 °}{\tan 36 °} + \frac{\tan 20 °}{\cot 70 °} =$

(2)

$\frac{\cot 54 °}{\tan 36 °} + \frac{\tan 20 °}{\cot 70 °}$

$= \frac{\cot (90 ° - 36 °)}{\tan 36 °} + \frac{\tan (90 ° - 70 °)}{\cot 70 °}$

$= \frac{\tan 36 °}{\tan 36 °} + \frac{\cot 70 °}{\cot 70 °} = 1 + 1 = 2$

Q 25 :

If $\frac{x}{\cos θ} = \frac{y}{\cos [θ - (\frac{2 π}{3})]} = \frac{z}{\cos [θ + (\frac{2 π}{3})]},$ then $x + y + z$ is equal to

(0)

We have $\frac{x}{\cos θ} = \frac{y}{\cos [θ - (\frac{2 π}{3})]} = \frac{z}{\cos [θ + (\frac{2 π}{3})]}$

Therefore, each ratio is equal to

$\frac{x + y + z}{\cos θ + \cos [θ - (\frac{2 π}{3})] + \cos [θ + (\frac{2 π}{3})]}$

$= \frac{x + y + z}{\cos θ + 2 \cos θ \cos (\frac{2 π}{3})}$

$= \frac{x + y + z}{0}$

$\Rightarrow x + y + z = 0$

Q 26 :

If $0 ° < θ < 180 °$ , then $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (1 + \cos θ)}}}}$ there being $n$ number of 2's, is equal to

$2 \cos \frac{θ}{2^{n}}$
$2 \cos \frac{θ}{2^{n - 1}}$
$2 \cos \frac{θ}{2^{n + 1}}$
none of these

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(1)

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (1 + \cos θ)}}}},$ there being $n$ numbers of 2's

$= \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (2 \cos^{2} \frac{θ}{2})}}}}$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (1 + \cos \frac{θ}{2})}}}},$

there being $(n - 1)$ number of 2's,

$= \dots = \sqrt{2 (1 + \cos \frac{θ}{2^{n - 1}})} = 2 \cos \frac{θ}{2^{n}}$

Q 27 :

If $cosec θ - \cot θ = p,$ then the value of $cosec θ$ is

$\frac{1}{2} (p + \frac{1}{p})$
$\frac{1}{2} (1 - \frac{1}{p})$
$(1 + \frac{1}{p})$
$(1 - \frac{1}{p})$

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(1)

Since $cosec θ - \cot θ = p$

$\Rightarrow$ $cosec θ + \cot θ = \frac{1}{p}$

Adding (i) and (ii), we have

$2 cosec θ = p + \frac{1}{p}$

$∴$ $cosec θ = \frac{1}{2} (p + \frac{1}{p})$

Q 28 :

$\tan 3 x - \tan 2 x - \tan x$ is equal to

$\tan x \tan 2 x \tan 3 x$
$- \tan x \tan 2 x \tan 3 x$
$\tan x \tan 2 x - \tan x \tan 3 x - \tan 2 x \tan 3 x$
none of these

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(1)

$\tan 3 x = \tan (2 x + x) = \frac{\tan 2 x + \tan x}{1 - \tan 2 x \tan x}$

$∴$ $\tan 3 x - \tan 3 x \tan 2 x \tan x = \tan 2 x + \tan x$

$∴$ $\tan 3 x - \tan 2 x - \tan x = \tan 3 x \tan 2 x \tan x$

Topic Question Set

Q 11 :

The value of $\sin 12 ° \sin 48 ° \sin 54 °$ is equal to

Q 12 :

The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is

Q 13 :

The maximum value of $\cos^{2} (\frac{π}{3} - x) - \cos^{2} (\frac{π}{3} + x)$ is

Q 14 :

If sum of all the solution of the equation $8 \cos x (\cos (\frac{π}{6} + x) \cdot \cos (\frac{π}{6} - x) - \frac{1}{2}) = 1$ in $[0, π]$ is $k π$ , then $k$ is equal to

Q 15 :

If $\sin^{2} θ = \frac{4 x y}{{(x + y)}^{2}}$ is true, if and only if

Q 16 :

If $A = \sin^{2} x + \cos^{4} x,$ then for all real $x$

Q 17 :

If $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x),$ where $x \in R, k \geq 1$ , then $f_{4} (x) - f_{6} (x)$ is equal to

Q 18 :

For $0 < ϕ < \frac{π}{2}$ , if $x = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ, y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ \sin^{2 n} ϕ,$ then

Q 19 :

The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

Q 20 :

The angles of a triangle are in A.P. The number of degrees in the least is to be the number of radians in the greatest as $60 : π$ . Then the greatest angle is

Q 21 :

The value of $6 (\sin^{6} θ + \cos^{6} θ) - 9 (\sin^{4} θ + \cos^{4} θ) + 4$ is

Q 22 :

The value of $\tan 15 ° + \cot 15 °$ is

Q 23 :

If $\sin θ_{1} + \sin θ_{2} + \sin θ_{3} = 3,$ then $\cos θ_{1} + \cos θ_{2} + \cos θ_{3} =$

Q 24 :

The value of $\frac{\cot 54 °}{\tan 36 °} + \frac{\tan 20 °}{\cot 70 °} =$

Q 25 :

If $\frac{x}{\cos θ} = \frac{y}{\cos [θ - (\frac{2 π}{3})]} = \frac{z}{\cos [θ + (\frac{2 π}{3})]},$ then $x + y + z$ is equal to

Q 26 :

If $0 ° < θ < 180 °$ , then $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (1 + \cos θ)}}}}$ there being $n$ number of 2's, is equal to

Q 27 :

If $cosec θ - \cot θ = p,$ then the value of $cosec θ$ is

Q 28 :

$\tan 3 x - \tan 2 x - \tan x$ is equal to

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Topic Question Set

Q 11 : The value of sin12°sin48°sin54° is equal to

Q 12 : The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is

Q 13 : The maximum value of cos2(π3-x)-cos2(π3+x) is

Q 14 : If sum of all the solution of the equation 8cosx(cos(π6+x)·cos(π6-x)-12)=1 in [0,π] is kπ, then k is equal to

Q 15 : If sin2θ=4xy(x+y)2 is true, if and only if

Q 16 : If A=sin2x+cos4x, then for all real x

Q 17 : If fk(x)=1k(sinkx+coskx), where x∈R, k≥1, then f4(x)-f6(x) is equal to

Q 18 : For 0<ϕ<π2, if x=∑n=0∞cos2nϕ, y=∑n=0∞sin2nϕ and z=∑n=0∞cos2nϕ sin2nϕ, then

Q 19 : The equation esinx-e-sinx-4=0 has

Q 20 : The angles of a triangle are in A.P. The number of degrees in the least is to be the number of radians in the greatest as 60:π. Then the greatest angle is

Q 21 : The value of 6(sin6θ+cos6θ)-9(sin4θ+cos4θ)+4 is

Q 22 : The value of tan15°+cot15° is

Q 23 : If sinθ1+sinθ2+sinθ3=3, then cosθ1+cosθ2+cosθ3=

Q 24 : The value of cot54°tan36°+tan20°cot70°=

Q 25 : If xcosθ=ycos[θ-(2π3)]=zcos[θ+(2π3)], then x+y+z is equal to

Q 26 : If 0°<θ<180°, then 2+2+2+⋯+2(1+cosθ) there being n number of 2's, is equal to

Q 27 : If cosecθ-cotθ=p, then the value of cosecθ is

Q 28 : tan3x-tan2x-tanx is equal to

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Q 11 :

The value of $\sin 12 ° \sin 48 ° \sin 54 °$ is equal to

Q 12 :

The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is

Q 13 :

The maximum value of $\cos^{2} (\frac{π}{3} - x) - \cos^{2} (\frac{π}{3} + x)$ is

Q 14 :

If sum of all the solution of the equation $8 \cos x (\cos (\frac{π}{6} + x) \cdot \cos (\frac{π}{6} - x) - \frac{1}{2}) = 1$ in $[0, π]$ is $k π$ , then $k$ is equal to

Q 15 :

If $\sin^{2} θ = \frac{4 x y}{{(x + y)}^{2}}$ is true, if and only if

Q 16 :

If $A = \sin^{2} x + \cos^{4} x,$ then for all real $x$

Q 17 :

If $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x),$ where $x \in R, k \geq 1$ , then $f_{4} (x) - f_{6} (x)$ is equal to

Q 18 :

For $0 < ϕ < \frac{π}{2}$ , if $x = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ, y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ \sin^{2 n} ϕ,$ then

Q 19 :

The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

Q 20 :

The angles of a triangle are in A.P. The number of degrees in the least is to be the number of radians in the greatest as $60 : π$ . Then the greatest angle is

Q 21 :

The value of $6 (\sin^{6} θ + \cos^{6} θ) - 9 (\sin^{4} θ + \cos^{4} θ) + 4$ is

Q 22 :

The value of $\tan 15 ° + \cot 15 °$ is

Q 23 :

If $\sin θ_{1} + \sin θ_{2} + \sin θ_{3} = 3,$ then $\cos θ_{1} + \cos θ_{2} + \cos θ_{3} =$

Q 24 :

The value of $\frac{\cot 54 °}{\tan 36 °} + \frac{\tan 20 °}{\cot 70 °} =$

Q 25 :

If $\frac{x}{\cos θ} = \frac{y}{\cos [θ - (\frac{2 π}{3})]} = \frac{z}{\cos [θ + (\frac{2 π}{3})]},$ then $x + y + z$ is equal to

Q 26 :

If $0 ° < θ < 180 °$ , then $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 (1 + \cos θ)}}}}$ there being $n$ number of 2's, is equal to

Q 27 :

If $cosec θ - \cot θ = p,$ then the value of $cosec θ$ is

Q 28 :

$\tan 3 x - \tan 2 x - \tan x$ is equal to