Question

cosec(90°- x)sin(180°- x)cot(360°- x)%sec(180°+ x)tan(90°+ x)sin(-x)=1

Answer 1

Answer:

We know that,

We know that,

cosec (90° -x) = sec x

sin(180°-x) = sin x

cot(360°-x) = - cot x

sec(180°+x) = - sec x

cot(360°-x) = - cot x

sec(180°+x) = - sec x

tan(90°+x) = - cot x

sin(-x) = -sin x

$LHS =\frac{cosec(90\degree -x)sin(180\degree -x)cot(360\degree -x)}{sec(180\degree +x)tan(90\degree +x)sin(-x) }$

$= \frac{secx \cdot sinx \cdot (-cotx) }{(-secx)(-cotx)(-sinx)}$

$= \frac{secx \cdot sinx \cdot cotx}{secx \cdot cotx \cdot sinx}\\= 1 \\=RHS$

$Hence\: proved$

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