Establish identity csc⁡θ−1cot⁡θ=cot⁡θcsc⁡θ+1

Question

Establish identity  csc⁡θ−1cot⁡θ=cot⁡θcsc⁡θ+1

crocolylec · Accepted Answer

Given identity csc⁡θ−1cot⁡θ=cot⁡θcsc⁡θ+1  csc⁡θ−1cot⁡θ=cot⁡θcsc⁡θ+1  First from the left-hand side,  csc⁡θ−1cot⁡θ  Here use the trigonometry reciprocal identity, csc⁡θ=1sin⁡θ and cot⁡θ=1tan⁡θ  csc⁡θ−1cot⁡θ=1sin⁡θ−1cos⁡θsin⁡θ  =1−sin⁡θsin⁡θcos⁡θsin⁡θ  =1−sin⁡θcos⁡θ  Multiply (1+sin⁡θ) in numerator and denominator,  csc⁡θ−1cot⁡θ=(1−sin⁡θ)(1+sin⁡θ)cos⁡θ(1+sin⁡θ)  =1−sin2θcos⁡θ(1+sin⁡θ)  =cos2θcos⁡θ(1+sin⁡θ)  =cos⁡θ(1+sin⁡θ)  multiply 1sin⁡θ in numerator and denominator,  csc⁡θ−1cot⁡θ=cos⁡θ(1sin⁡θ)(1+sin⁡θ)(1sin⁡θ)  =1−sin2θcos⁡θ(1+sin⁡θ)  =cos2θcos⁡θ(1+sin⁡θ)  =cos⁡θ(1+sin⁡θ)  multiply 1sin⁡θ in numerator and denominator,

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Establish identity. \frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}

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