Question

Question is attached.Step by step explanation only!​

Answer 1

Given :

• tan∅/1-cot∅ + cot∅/1-tan∅ = sin∅/cos∅ ÷ (1- cos∅/sin∅ ) + cos∅/sin∅ ÷ (1 - sin∅/cos∅)

To Do :-

• To prove LHS = RHS

Proof :-

We know that ,

• sin∅/cos∅ = tan∅
• cos∅/sin∅ = cot∅

On using above two in LHS ,

• LHS = tan∅/1-cot∅ + cot∅/1-tan∅
• LHS = tan∅/1-cot∅ + cot∅/1-tan∅ = sin∅/cos∅ ÷ (1- cos∅/sin∅ ) + cos∅/sin∅ ÷ (1 - sin∅/cos∅)
• LHS = RHS

Hence proved !

Answer 2

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{\tan\theta}{ 1- \cot \theta} + \frac{\cot\theta}{1 - \tan\theta} = \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}}$

Recall that tanθ = sinθ / cosθ. Likewise, cotθ = cosθ / sinθ. Then by substitution:

$\displaystyle \begin{aligned} \frac{\tan\theta}{ 1- \cot \theta} + \frac{\cot\theta}{1 - \tan\theta} & = \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} \\ \\ & \stackrel{\checkmark}{=} \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} \end{aligned}$

Hence verified.