Question

(tan theta /sec theta - 1)-(sin theta /1+cos theta) =2 cot theta ​

Answer 1

GIVEN :

The equation is $\frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta$

TO PROVE :

The given equation $\frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta$ is true

SOLUTION :

From the given equation is true

That is prove that LHS = RHS

Now taking LHS

$\frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}$

By using the trignometric identity :

$secx=\frac{1}{cosx}$

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

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

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

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

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

By using the algebraic identity :

$(a+b)(a-b)-a^2-b^2$

$=\frac{sin\theta+sin\theta cos\theta-sin\theta+sin\theta cos\theta}{1^2-cos^2\theta}$

Adding the like terms

$=\frac{2sin\theta cos\theta}{sin^2\theta}$

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

By using the trignometric identity :

$cotx=\frac{cos}{sinx}$

$=2cot\theta$=RHS

∴ LHS = RHS

∴ the equation $\frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta$ is true

Hence proved.