Chapter EP, Problem 14.2.14EP

To determine

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

Answer to Problem 14.2.14EP

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

Explanation of Solution

Given:

  $\frac{\text{tanθ-sinθ}}{\text{secθ}}\text{=}\frac{{\text{sin}}^{\text{3}}\text{θ}}{\text{1+cosθ}}$ .

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:

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

Calculation:

  $\frac{\text{tanθ-sinθ}}{\text{secθ}}\text{=}\frac{{\text{sin}}^{\text{3}}\text{θ}}{\text{1+cosθ}}$ .

Taking L.H.S:

  $\Rightarrow \frac{\frac{\text{sinθ-cosθsinθ}}{\text{cosθ}}}{\text{secθ}}$ .

  $\Rightarrow \frac{\text{sinθ}\left(\text{1-cosθ}\right)}{1}$ .

Multiplying both numerator and denominator by $\text{1+cosθ}$ :

  $\Rightarrow \frac{\text{sinθ}\left(\text{1-cosθ}\right)\left(1+\cos \text{θ}\right)}{1+\cos \text{θ}}$

  $\Rightarrow \frac{\text{sinθ}\left({\text{1-cos}}^2\text{θ}\right)}{1+\cos \text{θ}}$ .

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

  $\Rightarrow \frac{{\text{sin}}^3\text{θ}}{1+\cos \text{θ}}$ .

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