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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.

Answer.          [True]             
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.

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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.

Answer.      [False]            
Solution.     According to question
                            
Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In  \bigtriangleupABC
 \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.

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

Answer.       [False]            
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.

Answer.      [False]            

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 \bigtriangleupECD
\tan \theta _{1}= \frac{perpendicular}{Base}= \frac{DC}{EC}= \frac{H}{\iota }
Find tan\theta in \bigtriangleupECB 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.

 

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

Answer.      [False]            
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.

 

 

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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.

 

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

Answer.     [False]
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

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