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

Given:

$cos θ=\frac{12}{13}$

To do:

We have to show that $sin θ (1 – tan θ)=\frac{35}{156}$.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $cos\ \theta = cos\ A=\frac{12}{13}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (13)^2=(12)^2+BC^2$

$\Rightarrow BC^2=169-144$

$\Rightarrow BC=\sqrt{25}=5$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{5}{13}$

$tan\ \theta=\frac{BC}{AB}=\frac{5}{12}$

This implies,

Let us consider LHS,

$sin\ \theta(1-tan\ \theta)=\frac{5}{13}(1-\frac{5}{12})$

$=\frac{5}{13}(\frac{12-5}{12})$

$=\frac{5}{13} \times \frac{7}{12}$

$=\frac{35}{156}$

$=$ RHS

Hence proved. 

Tutorialspoint

Updated on: 10-Oct-2022

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