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#### If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top remains unchanged.

Solution.        According to question

In case-1. Height is H and observation distance is a.
In case-2, both height and observation distance is increased by 10%.
In case -1
$\tan \theta _{1}= \frac{H}{a}$         $\left ( \because \tan \theta = \frac{perpendicular}{Base} \right )$  .....(1)
In case -2
$\tan \theta _{2}= \frac{H+\frac{H}{10}}{a+\frac{a}{10}}$
$= \frac{\frac{11H}{10}}{\frac{11a}{10}}= \frac{11H}{10}\times \frac{10}{11a}= \frac{H}{a}$
$\tan \theta _{2}= \frac{H}{a} \cdots \left ( 2 \right )$
from equation (1) and (2) we observe that $\theta _{1}= \theta _{2}$
Hence the given statement is true.

#### The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

Solution.     According to question

Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In
$\bigtriangleup$
ABC

$\tan\theta=\frac{Perpendicular}{Base}$

$\tan 30^{\circ}= \frac{H}{a}$           $\left ( \because \theta = 30^{\circ} \right )$

$\frac{1}{\sqrt{3}}= \frac{H}{a}\; \cdots \left ( 1 \right )$       $\left ( \because \tan 30^{\circ} = \frac{1}{\sqrt{3}}\right )$

Case :2    When height is doubled

Here ED = a
In $\bigtriangleup DEF$
$\tan \theta = \frac{2H}{a}$
$\tan \theta = \frac{2}{3}$   (from (1))
But $\tan 60^{\circ}= \sqrt{3}$         (If the angle is double)
$\sqrt{3}\neq \frac{2}{\sqrt{3}}$
hence the given statement is false.

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#### $\cos \theta = \frac{a^{2}+b^{2}}{2ab}$ where a and b are two distinct numbers such that ab> 0.

We know that

$-1\leq \cos \theta \leq 1$
We also know that

$\left ( a-b \right )^{2}= a^{2}+b^{2}-2ab$
$\text{Since }$  $\left ( a-b \right )^{2}$ $\text{is a square term hence it is always positive }$
$\left ( a-b \right )^{2}> 0$
a2 + b2 – 2ab > 0
$a^{2}+b^{2}> 2ab$
We observe that $a^{2}+b^{2}$ is always greater than 2ab.
Hence,

$\frac{a^{2}+b^{2}}{2ab}> 1$
Because if we divide a big term by small then the result is always greater than 1.
cos$\theta$ is always less than or equal to 1
Hence the given statement is false.

#### The value of 2sinθ can be a+1/a , where a is a positive number, and a ≠ 1.

Solution.    We know that
-1≤ sin θ ≤ 1
Multiply by 2.
-2≤ 2 sin θ ≤ 2
Here we found that value of 2 sin $\theta$ is lies from – 2 to 2.
But if we take a > 0 and a $\neq$ 1 then
$a+\frac{1}{a}> 2$
For example a = 3
3 + 1/3 = 3.33
Hence $a+\frac{1}{a}$ is always greater than 2 in case of positive number except 1
But value of 2 sin $\theta$  is not greater than 2
Hence the given statement is false

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#### If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Solution.      According to question.

In the figure $\theta _{1}$ is the angle of elevation and $\theta _{2}$ is the angle of depression of the cloud.
For finding $\theta _{1}$find tanq in $\bigtriangleup$ECD
$\tan \theta _{1}= \frac{perpendicular}{Base}= \frac{DC}{EC}= \frac{H}{\iota }$
Find tan$\theta$ in $\bigtriangleup$ECB for $\theta _{2}$
$\tan \theta _{2}= \frac{BC}{EC}= \frac{3}{\iota }$
Here we found that $\theta _{1}$ and $\theta _{2}$ both are different.
Hence the given statement is false.

#### If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

Solution.     Let us take 2 cases.

Case 1:

Case 2 :

$\theta _{1}$ is the angle when length is small and $\theta _{2}$ is the angle when shadow length is increased.
for finding $\theta _{1}$, $\theta _{2}$ find tan $\theta$

$\tan \theta_{1} = \frac{Perpendicular}{Base}= \frac{height\, of\, tower}{length\, of\, shadow}$
$\tan \theta_{2} = \frac{height\, of\, tower}{length\, of\, shadow}$
In both the case height of the tower is the same but in case 2 length of the shadow is increased and if the length of shadow increased value of $\theta _{2}$ decreased.
Hence the given statement is false.

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#### (tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ.

(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ
Taking L.H.S.
(tanθ + 2) (2 tan θ + 1)
tanθ.(2tan θ+1) + 2(2tan θ +1)
$2\tan ^{2}\theta +\tan \theta +4\tan \theta+2$
$2\tan ^{2}\theta +5\tan \theta+2$
$2\left ( \tan ^{2}\theta +1 \right )+5\tan \theta\: \cdots \left ( 1 \right )$
We know that

$\sec ^{2 }\theta-\tan ^{2}\theta= 1$
$\left ( 1+\tan ^{2}\theta= \sec ^{2}\theta \right )$

Put the above value in (1)we get
$2\sec ^{2}\theta +5\tan \theta \neq 5\tan \theta +\sec ^{2}\theta$
L.H.S. $\neq$ R.H.S.
Hence the given expression is false.

#### If cosA + cos2A = 1, then sin2A + sin4A = 1.

Given
cosA + cos2A = 1        …(1)
cos A = 1 – cos2A
cosA = sin2A          …(2)    ($\because$ sin2 $\theta$ = 1 – cos2$\theta$)
$\sin ^{2} A+\sin ^{4}A= 1$
L.H.S.

$\sin ^{2} A+\sin ^{4}A$
$\sin ^{2} A+ \left ( \sin ^{2}A \right )^{2}$
$\cos A+\left ( \cos A \right )^{2}$ (from (2))
cosA + cos2A
= 1       (R.H.S.)           (from (1))
Hence sin2A + sin4A = 1
So, the given statement is true.

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#### Prove that $\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta }= \tan \theta$

L.H.S

$\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta } \cdots \left ( 1 \right )$
We know that
$1-\cos ^{2}\theta = \sin ^{2}\theta\text{ in (1)}$
$\sqrt{\sin ^{2}\theta \cdot \sec ^{2}\theta }$
$=\sqrt{\sin ^{2}\theta \times \frac{1}{\cos ^{2}\theta } }$             $\left ( \because \sec \theta = \frac{1}{\cos \theta } \right )$
$=\sqrt{\frac{\sin ^{2}\theta }{\cos ^{2}\theta }}$
$=\sqrt{\tan ^{2}\theta }= \tan \theta$                   $\left ( \because \frac{\sin \theta }{\cos \theta }= \tan \theta \right )$

L.H.S. = R.H.S.

Hence the given expression is true.

#### The value of the expression (sin 80° – cos80°) is negative.

Solution.     (sin 80° – cos 80°)
We know that from 0 to 90° sin$\theta$  and cos$\theta$  both are positive i.e.
$0< \sin \theta \leq 90^{\circ}$  (always positive)
$0< \cos \theta \leq 90^{\circ}$ (always positive)
At 45° both the values of sin$\theta$  and cos$\theta$  are the same but after 45° to 90° value of sin is greater than the value of cos$\theta$ Hence sin 80° > cos 80°.
If we subtract a smaller term from bigger than the result is positive.
Hence (sin 80° – cos80°) > 0
So, the given statement is false