If cos θ=1213, show that sin θ (1 − tan θ)=35156.

Solution:

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

$\sin \theta(1-\tan \theta)=\frac{35}{156}$