Question

iv)
cosec (90° - x) sin(180°- x)cot(360° - x)
sec(180° + x) tan (90° + x) sin(-x)
=​

Answer 1

Solution:-

$\rm \implies \csc(90^{o} -x)\sin(180^o-x)\cot(360^o-x)\sec(180^o+x)\tan(90^o+x)\sin(-x)$

Now we can write in radian form

$\rm \implies \csc(\dfrac{\pi}{2} -x)\sin(\pi-x)\cot(2\pi-x)\sec(\pi+x)\tan(\dfrac{\pi}{2}+x)\sin(-x)$

Now  using reduction of angle concept

$\to\rm\csc(\dfrac{\pi}{2}-x )=\sec x$

$\rm\to\sin(\pi-x)=sin x$

$\rm\to\cot(2\pi-x)=-\cot x$

$\rm\to\sec(\pi+x)=-\sec x$

$\rm\to\tan(\dfrac{\pi}{2}+x )= -\cot x$

$\rm\to\sin(-x)=-\sin x$

Now put the value

$\rm\implies \sec x\: \times\:\sin x\: \times\: -\cot x\:\times\:-\sec x\:\times\:-\cot x\:\times\:-\sin x$

Using trigonometry identities

$\rm\to\sec x=\dfrac{1}{\cos x}$

$\rm\to\cot=\dfrac{\cos x}{\sin x}$

Now we get

$\rm\implies \dfrac{1}{\cos x} \times\sin x\times\dfrac{\cos x}{\sin x} \times\dfrac{1}{\cos x} x\times\dfrac{\cos x}{\sin x}\times\sin x$

$\rm\implies \dfrac{1}{\cancel{\cos x}} \times\sin x\times\dfrac{\cancel{\cos x}}{\sin x} \times\dfrac{1}{\cancel{\cos x}} x\times\dfrac{\cancel{\cos x}}{\sin x}\times\sin x$

$\rm\implies\sin x\times\dfrac{1}{\sin x} \times\dfrac{1}{\sin x}\times\sin x$

Answer= 1