Question

find value of (sin 55- cos 55)/sin10

Answer 1

Answer:

Step-by-step explanation:

(sin55-cos55)/sin10

sin(90-35)-cos55/sin10

cos35-cos55/sin10

[-2sin(35+55/2).sin(35-55/2)]/sin10

[+2sin45.sin10]/sin10

/sqrt{2}

Answer 2

Answer:

$\frac{\sin 55^\circ-\cos 55^\circ}{\sin 10^\circ}=\sqrt2$

Step-by-step explanation:

Given : Expression $\frac{\sin 55^\circ-\cos 55^\circ}{\sin 10^\circ}$

To find : The value of the expression?

Solution :

Expression $\frac{\sin 55^\circ-\cos 55^\circ}{\sin 10^\circ}$

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

$\sin 55=\cos(90-55)$

$\sin 55=\cos(35)$

Substitute in the expression,

$=\frac{\cos 35^\circ-\cos 55^\circ}{\sin 10^\circ}$

Applying identity, $\cos A-\cos B=-2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$

$\cos 35-\cos 55=-2\sin(\frac{35+55}{2})\sin(\frac{35-55}{2})$

$\cos 35-\cos 55=-2\sin(45)\sin(-10)$

We know, $\sin(-\theta)=-\sin\theta$

$\cos 35-\cos 55=2\sin(45)\sin(10)$

Substitute in expression,

$=\frac{2\sin(45)\sin(10)}{\sin 10}$

$=2\sin(45)$

$=2\times\frac{1}{\sqrt2}$

$=\sqrt2$

Therefore, $\frac{\sin 55^\circ-\cos 55^\circ}{\sin 10^\circ}=\sqrt2$