Question

If $cos\Theta =\frac{12}{13}$, show that $sin\Theta(1-tan\Theta)=\frac{35}{156}$.

Answer 1

SOLUTION IS IN THE ATTACHMENT.

** Trigonometry is the study of the relationship between the sides and angles of a triangle.

The ratio of the sides of a right angled triangle with respect to its acute angles are called trigonometric ratios.

** For any acute angle in a right angle triangle the side opposite to the acute angle is called a perpendicular(P),  the side adjacent to this acute angle is called the base(B) and side opposite to the right angle is called the hypotenuse(H).

** Find the third  side of the right ∆ ABC by using Pythagoras theorem (AC² = AB² + BC²).

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Given,

$\cos\theta=\frac{12}{13}$

We know  that,

$\sin\theta=\sqrt{1-\cos^{2}\theta}\\\;\\\sin\theta=\sqrt{1-(\frac{12}{13})^{2}}\\\;\\\sin\theta=\sqrt{\frac{13^{2}-12^{2}}{13^{2}}}\\\;\\\sin\theta=\sqrt{\frac{169-144}{13^{2}}}\\\;\\\sin\theta=\sqrt{\frac{25}{13^{2}}}\\\;\\\sin\theta=\frac{5}{13}$

Again,

$\tan\theta=\frac{\sin\theta}{\cos\theta}\\\;\\\tan\theta=\frac{\frac{5}{13}}{\frac{12}{13}}\\\;\\\tan\theta=\frac{5}{12}$

Now,

$\sin\theta(1-\tan\theta)=\frac{5}{13}(1-\frac{5}{12})\\\;\\\sin\theta(1-\tan\theta)=\frac{5}{13}(\frac{12-5}{12})\\\;\\\sin\theta(1-\tan\theta)=\frac{5}{13}\times\frac{7}{12}\\\;\\\sin\theta(1-\tan\theta)=\frac{35}{156}$