Question

tan theta/1-tantheta -cottheta/1-cottheta = costheta + sintheta/ costheta-sinthetaa

Answer 1

Please refer to the attachment for your answer .

Answer 2

Given that the  trignometric equation is $\frac{tan\theta}{1-tan\theta}-\frac{cot\theta}{1-cot\theta}=\frac{cos\theta+sin\theta}{cos\theta-sin\theta}$

To prove the equivalent property for the given trignometric equation :

Solution:

That is to prove that LHS=RHS

Now taking the LHS:

Simplifying the expression as below :

$\frac{tan\theta}{1-tan\theta}-\frac{cot\theta}{1-cot\theta}$

Here by using the trignometric identities:

( $tanx=\frac{sinx}{cosx}$ and $cotx=\frac{cosx}{sinx}$ )

$=\frac{\frac{sin\theta}{cos\theta}}{1-\frac{sin\theta}{cos\theta}}-\frac{\frac{cos\theta}{sin\theta}}{1-\frac{cos\theta}{sin\theta}}$

$=\frac{\frac{sin\theta}{cos\theta}}{\frac{cos\theta-sin\theta}{cos\theta}}-\frac{\frac{cos\theta}{sin\theta}}{\frac{sin\theta-cos\theta}{sin\theta}}$

$=\frac{sin\theta}{cos\theta}\times \frac{cos\theta}{cos\theta-sin\theta}-\frac{cos\theta}{sin\theta}\times \frac{sin\theta}{sin\theta-cos\theta}$

$=\frac{sin\theta}{cos\theta-sin\theta}-\frac{cos\theta}{sin\theta-cos\theta}$

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

$=\frac{sin\theta}{cos\theta-sin\theta}+\frac{cos\theta}{cos\theta-sin\theta}$

$=\frac{sin\theta+cos\theta}{cos\theta-sin\theta}$ ( by taking LCM )

Rewritting we get

$=\frac{cos\theta+sin\theta}{cos\theta-sin\theta}$

= RHS

Therefore we get that LHS=RHS

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

Hence the given trignometric equation $\frac{tan\theta}{1-tan\theta}-\frac{cot\theta}{1-cot\theta}=\frac{cos\theta+sin\theta}{cos\theta-sin\theta}$ is proved .