Chapter EP, Problem 14.2.13EP

To determine

To prove: the equation is in the form of identity.

Answer to Problem 14.2.13EP

The both L.H.S AND R.H.S are equal or identity to each other.

Explanation of Solution

Given:

  $\frac{{\cot }^{\text{2}}\text{θ}}{1+{\text{cot}}^{\text{2}}\text{θ}}{\text{=1-sin}}^{\text{2}}\text{θ}$ .

Concept used:

Trigonometric ratios are:

$\text{tanθ=}\frac{\text{sinθ}}{\text{cosθ}}$ and $\text{cotθ=}\frac{\text{cosθ}}{\text{sinθ}}$ .

$\text{cosecθ=}\frac{\text{1}}{\text{sinθ}}$ , $\text{secθ=}\frac{\text{1}}{\text{cosθ}}$ , $\text{cotθ=}\frac{\text{1}}{\text{tanθ}}$ .

Trigonometric identities:

  ${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$ .

Which implies:

  ${\text{1-sin}}^2{\text{θ=cos}}^2\text{θ}$ .

Calculation:

  $\frac{{\cot }^{\text{2}}\text{θ}}{1+{\text{cot}}^{\text{2}}\text{θ}}{\text{=1-sin}}^{\text{2}}\text{θ}$ .

  $\Rightarrow \frac{\frac{{\cos }^{\text{2}}\text{θ}}{{\sin }^{\text{2}}\text{θ}}}{1+\frac{{\cos }^{\text{2}}\text{θ}}{{\sin }^{\text{2}}\text{θ}}}{\text{=1-sin}}^{\text{2}}\text{θ}$ .

According to the trigonometric formula:

  ${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$ ;

  $\Rightarrow \frac{\frac{{\cos }^{\text{2}}\text{θ}}{{\sin }^{\text{2}}\text{θ}}}{\frac{{\sin }^{\text{2}}\text{θ}+{\cos }^{\text{2}}\text{θ}}{{\sin }^{\text{2}}\text{θ}}}{\text{=1-sin}}^{\text{2}}\text{θ}$ .

  $\Rightarrow {\cos }^{\text{2}}{\text{θ=1-sin}}^{\text{2}}\text{θ}$ .

If ${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$ then ${\text{cos}}^{\text{2}}{\text{θ=1-sin}}^{\text{2}}\text{θ}$ ;

  $\Rightarrow 1-{\sin }^{\text{2}}{\text{θ=1-sin}}^{\text{2}}\text{θ}$ .

So, the both L.H.S AND R.H.S are equal or identity to each other.