In Fig. $AD = 4cm, BD = 3cm$ and $CB = 12cm$, find the $\cot\theta.$
A$\frac{12}{5}$
B$\frac{5}{12} $
C$\frac{13}{12}$
D$\frac{12}{13}$
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A$\frac{12}{5}$
We have the following given data in the figure, AD= 4cm, BD = 3cm, CB = 12cm
Now we will use Pythagoras theorem in $\triangle\text{ABD},$
$\text{AB}=\sqrt{3^2+4^2}$
$=5\text{cm}$
Therefore,
$\cot\theta=\frac{\text{CB}}{\text{AB}}$
$=\frac{12}{5}$
So the answer is $(a)$
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If $\theta$ is an acute angle such that $\tan^2\theta=\frac{8}{7},$ then the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ is: