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#### A pole 6 m high casts a shadow 2$\sqrt{3}$ m long on the ground, then the Sun’s elevation is(A) 60°            (B) 45°            (C) 30°            (D) 90°

Solution.         Given :
height pole = 6 m
Shadow of pole = $2\sqrt{3}m$
Now make figure according to given condition

Let angle of elevation is $\alpha$
$\therefore \tan \alpha = \frac{Perpendicular}{base}$
$\tan \alpha = \frac{6}{2\sqrt{3}}$
$\tan \alpha = \frac{3}{\sqrt{3}}= \sqrt{3}$
$\tan \alpha = \tan 60^{\circ}$              $\left [ \because \tan 60^{\circ}=\sqrt{3} \right ]$
a = 60°
Hence the Sun's elevation is 60°.

#### sin (45° + θ) – cos (45° – θ) is equal to (A) 2cosθ        (B) 0                (C) 2sinθ         (D) 1

Solution.    Here

:$\sin \left ( 45^{\circ}+\theta \right )-\cos \left ( 45^{\circ}-\theta \right )$
Sin[90° - (45°- θ)] – cos(45°- θ)
$\left [ \because \left ( 45^{\circ} +\theta \right ) = \left ( 90^{\circ}-\left ( 45-\theta \right ) \right )\right ]$
Cos(45°- θ) – cos(45°- θ)         [$\because$ sin (90 – θ) = cosθ]
= 0
Hence option (B) is correct

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#### If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is(A) 1                (B)3/4                     (C) 1/2                   (D) 1/4

$\sin\theta-\cos\theta=0$
squaring both sides we get
$(\sin\theta-\cos\theta)^2=0$
$\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=0$
($\therefore$ (a – b)2 = a2 + b2 – 2ab)
$\sin^2\theta+\cos^2\theta=2\sin\theta\cos\theta.......(1)$
$1=2\sin\theta\cos\theta\because (\sin^2\theta+\cos^2\theta=1)$
$\frac{1}{2}=\sin \theta \cos \theta$
Squaring both sides we get

$\frac{1}{4}=\sin^{2} \theta \cos^{2} \theta$       …(2)
Now squaring both side of equation (1) we get
$\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{2}= \left ( 2\sin \theta \cos \theta \right )^{2}$
$\left ( \sin ^{2}\theta \right )^{2}+\left ( \cos ^{2}\theta \right )^{2}+2\sin ^{2}\theta \cdot \cos ^{2}\theta = 4\sin ^{2}\theta \cos ^{2}\theta$
$\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right ]$
$\left ( \sin ^{4}\theta \right )+\left ( \cos ^{4}\theta \right )= 4\sin ^{2}\theta \cos ^{2}\theta -2\sin ^{2}\theta \cos ^{2}\theta$
$\sin ^{4}\theta +\cos ^{4}\theta = 2\sin ^{2}\theta \cos ^{2}\theta$
(Use equation (2))
$\sin ^{4}\theta +\cos ^{4}\theta = 2\left ( \frac{1}{4} \right )$
$\sin ^{4}\theta +\cos ^{4}\theta =\frac{1}{2}$
Hence option (C) is correct.

#### If  $4 \tan\theta= 3$ then  $\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )$  is equal to$(A)\frac{2}{3}$                     $(B)\frac{1}{3}$                     $(C)\frac{1}{2}$                $(D)\frac{3}{4}$

Hence option (C) is correct.

$\\\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )\text{ Divide numerator and denominator by and rewriting the given expression}\\\\\Rightarrow ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta })=\frac{4\frac{\sin\theta}{cos\theta}-\frac{cos\theta}{cos\theta}}{4\frac{\sin\theta}{cos\theta}+\frac{cos\theta}{cos\theta}}\\\\=\frac{4\tan\theta-1}{4\tan\theta+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\ (\text{Given }4\tan\theta=3)$

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#### The value of the expression $\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]\ is$(A) 3                (B) 2                (C) 1                (D) 0

Solution.

$\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin \left ( 90^{\circ}-63^{\circ} \right )$
$\left [ \because \sin \left ( 90^{\circ}-\theta \right )= \cos \theta \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos 63^{\circ}\times \cos 63^{\circ}$
$= \frac{\sin ^{2}22^{\circ}+\sin ^{2}\left (90^{\circ}-22^{\circ} \right )}{\cos ^{2}\left ( 90^{\circ}-68^{\circ} \right )+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because 68^{\circ}= \left ( 90^{\circ}-22^{\circ} \right ) & \\ 22^{\circ}= \left ( 90^{\circ}-68^{\circ} \right )& \end{Bmatrix}$
$= \frac{\sin ^{2}22^{\circ}+\cos ^{2}22^{\circ}}{\sin ^{2}68^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta & \\ \cos \left ( 90^{\circ} -\theta \right )= \sin \theta & \end{Bmatrix}$
$= \frac{1}{1}+1\; \left [ \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right ]$
= 1+ 1 =2
Hence option (B) is correct.

#### Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is(A) 0°              (B) 30°            (C) 60°            (D) 90°

Solution.

$\\\sin \alpha=1/2\Rightarrow \alpha=30^0\\\cos \beta=1/2\Rightarrow \beta=60^0\\\therefore \alpha+\beta=90^0$
Hence option (D) is correct.

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#### If sinA + sin2A = 1, then the value of the expression (cos2A+ cos4A) is(A) 1                (B)1/2                (C) 2                (D) 3

Solution.   It is given that sinA + sin2A = 1          …(*)
sinA = 1 – sin2A
sinA = cos2A               …(1)       ( $\because$ 1 – sin2A = cos2A)
Squaring both sides we get
sin2 A = cos4A             …(2)
Hence cos2A + cos4A =
= sinA + sin2A            {using (1) and (2)}
= sinA + sin2A = 1      (given)
Hence option (A) is correct.

#### If ΔABC is right angled at C, then the value of cos (A+B) is$(A) 0 \ \ \ \ \ (B) 1 \ \ \ \ \ (C)1/2 \ \ \ \ \ (D)\frac{\sqrt{3}}{2}$

Solution.     It is given that $\angle$C = 90°

In $\bigtriangleup$ABC

$\angle$A +$\angle$B +$\angle$C = 180             [$\because$  sum of interior angles of triangle is 180°]
$\angle$A + $\angle$B + 90o = 180            [$\because$ C = 90° (given)]
$\angle$A + $\angle$B = 1800 – 900
$\angle$A + $\angle$B = 90°          …(1)
cos($\angle$A +$\angle$B) = cos (90°)
cos(90°) = 0
[$\because$ from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.

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#### If cos 9α = sinα and 9α < 90° , then the value of tan5α is   (A) $\frac{1}{\sqrt{3}}$            (B) ${\sqrt{3}$             (C) 1                (D) 0

Solution.         Given :- cos 9$\alpha$ = sin$\alpha$
cos9 $\alpha$ = cos(90 – $\alpha$)
$\because$ (cos (90 – $\alpha$) = sin$\alpha$)
9$\alpha$ = 90 – $\alpha$
9$\alpha$+ $\alpha$ = 90
10 $\alpha$= 90
$\alpha = \frac{90}{10}$
Now tan 5$\alpha$ is
Put $\alpha$ = 9 we get
tan 5$\times$ (9)
tan 450
= 1
{$\because$ from the table of trigonometric ratios of angles we know that tan 450 = 1}
Hence option C is correct.

#### The value of (tan1° tan2° tan3° ... tan89°) is (A) 0                (B) 1                (C) 2                (D)1/2

Solution.         Given :-tan1° tan2° tan3° ... tan89°
tan1° tan2° tan3° ... tan89°tan87° tan 88° tan89°       …(1)
We can also write equation (1) in the form of
[tan (900 – 890) . tan (900 – 880). tan (900 – 870) …… tan 87° . tan 88° tan 89°]
[$\because$ we can write tan 1° in the form of tan (900 – 890) similarly we can write other values]
[cot 890 . cot 880. cot 870 …. tan 870 . tan 880 . tan 890]
$\because$   [tan (902$\theta$) = cot$\theta$ ]

Also

$\left [ \frac{1}{\tan 89^{\circ}}\frac{1}{\tan 88^{\circ}} \frac{1}{\tan 87^{\circ}}\cdots \tan 87^{\circ}.\tan 88^{\circ}\tan 89^{\circ} \right ]$

$\because$ Throughout all terms are cancelled by each other and remaining will be tan45

Hence the value is 1

$\because$ option B is correct.