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  • If cot Theta = 7 / 8 , evaluate : (1+\sin \Theta)(1-\sin \Theta) / (1+\cos \Theta)(1-\cos \Theta)

7.      If \cot \theta =\frac{7}{8}, evaluate :

            (i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
            (ii)\; \cot ^{2}\theta

             

Answers (1)

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Given that,
\cot \theta =\frac{7}{8}
\therefore perpendicular  (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which \angle B =90^0
Now, By using Pythagoras theorem,
AC^2 = AB^2+BC^2
AC = \sqrt{64 +49} =\sqrt{113}

So,        \sin \theta = \frac{BC}{AC} = \frac{8}{\sqrt{113}}       
and       \cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}

\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}
(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta
 =(\frac{7}{8})^2 = \frac{49}{64}

(ii)\; \cot ^{2}\theta 
  =(\frac{7}{8})^2 = \frac{49}{64}

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manish

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