Question

Sin 18°+ tan 72° - cos 72°/ cosec 75° + cot 18°- sec 15° = 1/2

Answer 1

Answer:

$\frac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}=1$

Step-by-step explanation:

Given : Expression $\frac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}$

To find : Solve the expression ?

Solution :

$\frac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}$

We can write it as,

$=\frac{\sin 18^\circ+\tan 72^\circ-\cos (90-18)^\circ}{\csc (90-15)^\circ+\cot 18^\circ-\sec 15^\circ}$

We know, $\cos(90-\theta)=\sin\theta$ and $\csc(90-\theta)=\sec\theta$

$=\frac{\sin 18^\circ+\tan 72^\circ-\sin 18^\circ}{\sec 15^\circ+\cot 18^\circ-\sec 15^\circ}$

$=\frac{\tan 72^\circ}{\cot 18^\circ}$

$=\frac{\tan (90-18)^\circ}{\cot 18^\circ}$

We know, $\tan(90-\theta)=\cot\theta$

$=\frac{\cot 18^\circ}{\cot 18^\circ}$

$=1$

Therefore, $\frac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}=1$

Answer 2

$\dfrac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}=1$

Step-by-step explanation:

In order to solve the provided expression use the identity:

$\cos(90-\theta)=\sin\theta, \\\tan(90-\theta)=\cot\theta\\ \csc(90-\theta)=\sec\theta$

Consider the provided equation.

$\dfrac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}$

The equation can be written as:

$=\dfrac{\sin 18^\circ+\tan (90-18)^\circ-\cos (90-18)^\circ}{\csc (90-15)^\circ+\cot 18^\circ-\sec 15^\circ}$

By using the identities.

$=\dfrac{\sin 18^\circ+\tan (90-18)^\circ-\cos (90-18)^\circ}{\csc (90-15)^\circ+\cot 18^\circ-\sec 15^\circ}$

$=\dfrac{\cot 18^\circ}{\cot 18^\circ}\\\\=1$

Therefore, $\dfrac{\sin 18^\circ+\tan 72^\circ-\cos 72^\circ}{\csc 75^\circ+\cot 18^\circ-\sec 15^\circ}=1$

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