Question

(tan theta)/(1-cot theta)+(cot theta)/(1-tan theta)=1+sec theta csc theta

Answer 1

please see the attachment

Answer 2

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \times \csc \theta$ Proved.

Step-by-step explanation:

Consider the provided information.

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \times \csc \theta$

Use the identity: $\begin{array}{l}{\tan \theta=\frac{\sin \theta}{\cos \theta}}, {\cot \theta=\frac{\cos \theta}{\sin \theta}}\end{array}$

Consider the LHS.

$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\left(\frac{\cos \theta}{\sin \theta}\right)}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\left(\frac{\sin \theta}{\cos \theta}\right)}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}\\=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$

$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \times \cos \theta(\sin \theta-\cos \theta)}$

Use the identity: $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$

$=\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta\right)}{\sin \theta \times \cos \theta(\sin \theta-\cos \theta)}$

Use the identity: sin²θ+cos²θ=1

$=\frac{1+\sin \theta \cos \theta}{\sin \theta \times \cos \theta}\\=\frac{1}{\sin \theta \times \cos \theta}+\frac{\sin \theta \cos \theta}{\sin \theta \times \cos \theta}=1+\sec \theta \times \csc \theta$

LHS=RHS

Hence, proved

#Learn more

Cos theta/1-tan theta + sin theta/1-cot theta

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