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  • Show that cos power 2 (45 degree + theta) + cos power 2 (45 degree - theta) by than (60 degree + theta)tan (30 degree - theta)= 1

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\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

Answers (1)

best_answer

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.

Posted by

infoexpert27

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