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Show that tan4θ + tan2θ = sec4θ – sec2θ.       

Taking L.H.S.
tan4θ + tan2θ
(tan2θ) + tan2θ…(1)
We know that sec2θ – tan2θ = 1
Put

\tan ^{2}\theta = \sec ^{2}\theta -1    in (1)
= \left ( \sec ^{2}\theta -1 \right )^{2}+\sec ^{2}\theta -1
 = \left ( \sec ^{2}\theta \right )^{2}+\left ( 1 \right )^{2}-2\left ( \sec ^{2}\theta \right )\left ( 1 \right )+\sec ^{2}\theta -1
\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )
= \sec ^{4}\theta +1-2\sec ^{2}\theta +\sec ^{2}\theta -1
 = \sec ^{4}\theta -\sec ^{2}\theta
  LHS = RHS
Hence proved        

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An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer.       [45°]               
Solution.      According to the question.
           
In \bigtriangleupEDC         EC = 20.5 m
DC = 20.5 m
To find angle \theta in \bigtriangleupEDC we need to find tan\theta.
\tan \theta = \frac{P}{B}
\tan \theta = \frac{EC}{DC}
\tan \theta = \frac{20\cdot 5}{20\cdot 5}
\tan \theta = 1
\theta = 45^{\circ}                 \left ( \because \tan 45^{\circ}= 1 \right )
Hence the angle of elevation is 45°.

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Show that

\frac{\cos ^{2}\left ( 45^{\circ}+\theta \right )+\cos ^{2}\left ( 45^{\circ}-\theta \right )}{\tan \left ( 60^{\circ} +\theta \right )\tan \left ( 30 ^{\circ}+\theta \right )}= 1

L.H.S
              = \frac{\cos ^{2}\left ( 90-\left ( 45-\theta ^{\circ} \right ) \right )+\cos ^{2}\left ( 45-\theta ^{\circ} \right ) }{\tan \left ( 60^{\circ}+\theta \right )\tan \left ( 30^{\circ}-\theta \right )}         \left ( \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta \right )
               = \frac{\sin ^{2}\left ( 45^{\circ}-\theta \right )+\cos ^{2}\left ( 45^{\circ} -\theta\right )}{\tan \left ( 90^{\circ}-\left (30^{\circ} -\theta \right ) \right )\tan \left ( 30^{\circ}-\theta \right )}               \left ( \because \tan \left ( 90-\theta \right )= \cot \theta \right )
              = \frac{\sin ^{2}\left ( 45-\theta \right )+\cos ^{2}\left ( 45-\theta \right )}{\cot \left ( 30^{\circ}-\theta \right )\tan \left ( 30^{\circ}-\theta \right )}                            \left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )
             = \frac{1}{\frac{1}{\tan \left ( 30^{\circ}-\theta \right )}\times \tan \left ( 30-\theta \right )}                                       \left ( \because \cot \theta= \frac{1}{\tan \theta} \right )
             = \frac{1}{1}=1

L.H.S. = R.H.S.
Hence proved.

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If 2sin2θ – cos2θ = 2, then find the value of θ.

  2sin2θ – cos2θ = 2
2\left ( 1-\cos ^{2}\theta \right )-\cos ^{2}\theta = 2           \left ( \because \sin ^{2}\theta +\cos ^{2}\theta= 1 \right )
2-2\cos ^{2}\theta-\cos ^{2}\theta= 2
2-3\cos ^{2}\theta-2= 0
-3\cos ^{2}\theta= 0
\cos ^{2}\theta= 0
\cos \theta= 0
\theta= 90^{\circ}           \left ( \because \cos 90^{\circ}= 0 \right )

 Hence value of \theta is 90°

 

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Simplify (1+tan^2\theta)(1-sin\theta)(1+sin\theta)

(1+tan^2\theta)(1-sin\theta)(1+sin\theta)
= \left ( \sec ^{2}\theta \right )\left ( \left ( 1 \right )^{2} -\left ( \sin \theta \right )^{2}\right )      \left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )
= \left ( \sec ^{2}\theta \right )\left ( 1-\sin ^{2}\theta \right )
= \left ( \sec ^{2}\theta \right )\left ( \cos ^{2}\theta \right )                \left ( \because \sin ^{2}\theta + \cos ^{2}\theta= 1 \right )
= \frac{1}{ \cos ^{2}\theta}\times \cos ^{2}\theta                   \left ( \because \sec ^{2}\theta = \frac{1}{\cos \theta } \right )
= 1

 

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A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

Length of ladder = 15 m
The angle between wall and ladder = 60°
Let the height of wall = H

In \bigtriangleupABC  \angleC = 60°, \angleB = 90°
We know that
\angleA + \angleB + \angleC = 180°   (Sum of interior angles of a triangle is 180)
\angleA + 90 + 60 = 180°
\angleA = 30°
In \bigtriangleupABC      

\sin 30^{\circ} = \frac{H}{15}
H= \sin 30^{\circ} \times 15
H= \frac{1}{2} \times 15        \left ( \because \sin 30^{\circ} = \frac{1}{2}\right )
H= 7\cdot 5\, m
Hence the height of the wall is 7.5 m

 

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If \sqrt{3}  tan θ = 1, then find the value of sin2θ – cos2θ.

   Given :

\sqrt{3}\tan\theta=1  
\tan \theta = \frac{1}{\sqrt{3}}
\theta = 30^{\circ}       \left ( \because \tan30^{\circ}= \frac{1}{\sqrt{3}} \right )
\sin ^{2}\theta -\cos ^{2}\theta = \sin ^{2}30^{\circ}-\cos ^{2}30^{\circ}
(Because θ = 300)
= \left ( \frac{1}{2} \right )^{2}-\left ( \frac{\sqrt{3}}{2} \right )^{2}         \begin{bmatrix} \because \sin 30^{\circ}= \frac{1}{2} & \\ \cos 30^{\circ}= \frac{\sqrt{3}}{2}& \end{bmatrix}
= \frac{1}{4}-\frac{3}{4}

Taking L.C.M.
= \frac{1-3}{4}= \frac{-2}{4}
\sin ^{2}\theta -\cos ^{2}\theta = \frac{-1}{2}

 

 

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Find the angle of elevation of the sun when the shadow of a pole h metres high is \sqrt{3} h metres long.

Answer.       [30°]   
Solution.       According to question
                       
 Here BC is the height of the pole i.e. h meters and AB is the length of shadow i.e. \sqrt{3}h .
 For finding angle q we have to find tanq in \bigtriangleupABC
\tan \theta = \frac{Perpendicular}{Base}
= \tan \theta = \frac{h}{\sqrt{3}h}
\theta = 30^{\circ}        \left ( \because \tan 30^{\circ}= \frac{1}{\sqrt{3}} \right )
Hence angle of elevation is 30°.

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Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)

Solution. 
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
  Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ    (\because tan (90 – θ) = cot θ)
  =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }      \left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )
Taking L.C.M.
\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\sin \theta \cos \theta }
= \frac{1}{\cos \theta \cdot \sin \theta }       \left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )
= \frac{1}{\cos \theta }\times \frac{1}{\sin \theta }
= \sec \theta \times \cos ec\, \theta     \left ( \because \frac{1}{\cos \theta } = \sec \theta ,\frac{1}{\sin \theta }= \cos ec\, \theta \right )
= \sec \theta \times \sec \left ( 90^{\circ}-\theta \right )          \left ( \because \sec \left ( 90-\theta \right ) = \cos ec\, \theta \right )
L.H.S. = R.H.S.
Hence proved.
 

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Prove the following :

1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha

1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha

Taking L.H.S.
           = 1+\frac{\cot ^{2}\alpha }{1+\cos ec\, \alpha }  
          
          = \frac{1+\cos ec\, \alpha +\cot ^{2}\alpha }{1+\cos ec\, \alpha }
        = \frac{\cos ec^{2}\alpha-\cot ^{2}\alpha+ \cos ec\, \alpha+\cot ^{2}\alpha \, \, }{1+\cos ec\, \alpha }      \left ( \because \cos ec^{2}\theta -\cot ^{2}\theta = 1 \right )
       = \frac{\cos ec^{2}\alpha +\cos ec\, \alpha }{1+\cos ec\, \alpha }
      = \frac{\cos ec\, \alpha \left ( \cos ec\, \alpha+1 \right ) }{\left ( 1+\cos ec\, \alpha \right )}
     = \cos ec\, \alpha
 L.H.S. = R.H.S.
Hence proved

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