If $\cos \theta=\frac{12}{13}$, show that $\sin \theta(1-\tan \theta)=\frac{35}{156}$
Given: $\cos \theta=\frac{12}{13}$....(1)
To show that $\sin \theta(1-\tan \theta)=\frac{35}{156}$
Now, we know that $\cos \theta=\frac{\text { Base side adjacent to } \angle \theta}{\text { Hypotenuse }}$....(2)
Therefore, by comparing equation (1) and (2)
We get,
Base side adjacent to $\angle \theta=12$
And
Hypotenuse = 13
Therefore from above figure
Base side $B C=12$
Hypotenuse $A C=13$
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
$A C^{2}=A B^{2}+B C^{2}$
Therefore by substituting the values of known sides
We get,
$13^{2}=A B^{2}+12^{2}$
Therefore,
$A B^{2}=13^{2}-12^{2}$
$A B^{2}=169-144$
$A B^{2}=25$
$A B=\sqrt{25}$
Therefore,
$A B=5 \ldots \ldots(3)$
Now, we know that
$\sin \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Hypotenuse }}$
Now from figure (a)
We get,
$\sin \theta=\frac{A B}{A C}$
Therefore,
$\sin \theta=\frac{5}{13}$..(4)
Now, we know that
$\tan \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Base side adjacent to } \angle \theta}$
Now from figure (a)
We get,
$\tan \theta=\frac{A B}{B C}$
Therefore,
$\tan \theta=\frac{5}{12}$..(5)
Now L.H.S. of the equation to be proved is as follows
L.H.S. $=\sin \theta(1-\tan \theta) \ldots \ldots(6)$
Substituting the value of $\sin \theta$ and $\tan \theta$ from equation (4) and (5) respectively
We get,
$L . H . S .=\frac{5}{13}\left(1-\frac{5}{12}\right)$
Taking L.C.M inside the bracket
We get,
L.H.S. $=\frac{5}{13}\left(\frac{1 \times 12}{1 \times 12}-\frac{5}{12}\right)$
L.H.S. $=\frac{5}{13}\left(\frac{7}{12}\right)$
Now, by opening the bracket and simplifying
We get,
$L . H . S .=\frac{5 \times 7}{13 \times 12}$
L.H.S. $=\frac{35}{156}$....(7)
From equation (6) and (7) , it can be shown that
$\sin \theta(1-\tan \theta)=\frac{35}{156}$