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) f) 1 + cot²a = cosec²a​

User Kakoma
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Answer:

It is an identity, proved below.

Explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).


\displaystyle \large{\cot x=(1)/(\tan x)}\\\displaystyle \large{\csc x=(1)/(\sin x)}

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.


\displaystyle \large{1+((1)/(\tan x))^2=((1)/(\sin x))^2}\\\displaystyle \large{1+(1)/(\tan^2x)=(1)/(\sin^2x)}

Another identity is:


\displaystyle \large{\tan x=(\sin x)/(\cos x)}

Therefore:


\displaystyle \large{1+(1)/(((\sin x)/(\cos x))^2)=(1)/(\sin^2x)}\\\displaystyle \large{1+(1)/((\sin^2x)/(\cos^2x))=(1)/(\sin^2x)}\\\displaystyle \large{1+(\cos^2x)/(\sin^2x)=(1)/(\sin^2x)}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.


\displaystyle \large{(\sin^2x)/(\sin^2x)+(\cos^2x)/(\sin^2x)=(1)/(\sin^2x)}\\\displaystyle \large{(\sin^2x+\cos^2x)/(\sin^2x)=(1)/(\sin^2x)

Another identity:


\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:


\displaystyle \large{(\sin^2x+\cos^2x)/(\sin^2x)=(1)/(\sin^2x)\longrightarrow \boxed{ (1)/(\sin^2x)={(1)/(\sin^2x)}}

Hence proved, this is proof by using identity helping to find the specific identity.

User Jskinner
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