Question

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

Answer 1

Step-by-step explanation:

which is required ans.a

Ans =1

(ans can't be 1)

Answer 2

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

Step-by-step explanation:

Consider the provided equation.

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

Use the identity: $\cos(90-\theta)=\sin\theta, \tan(90-\theta)=\cot\theta\ and\ \csc(90-\theta)=\sec\theta$

Rewrite the above equation as:

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

Use the above mentioned identity.

$=\frac{\sin 18^\circ+\cot18^\circ-\sin 18^\circ}{\sec 15^\circ+\cot 18^\circ-\sec 15^\circ}$

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

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

#Learn more

Base = 4 cm , Perpendicular = 3 cm. Find the value of all trigonometry ratio

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