Chapter EP, Problem 14.4.12EP

To determine

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

Answer to Problem 14.4.12EP

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

Explanation of Solution

Given:

  $\text{1+cos2θ=}\frac{\text{2}}{{\text{1+tan}}^{\text{2}}\text{θ}}$ .

Concept used:

Trigonometric formula:

  ${\text{cos2θ=2cos}}^{\text{2}}\text{θ-1}$ .

  ${\text{1+cos2θ=2cos}}^{\text{2}}\text{θ}$ .

  ${\text{1+tan}}^{\text{2}}{\text{θ=sec}}^{\text{2}}\text{θ}$ .

Calculation:

  $\text{1+cos2θ=}\frac{\text{2}}{{\text{1+tan}}^{\text{2}}\text{θ}}$ .

Taking L.H.S:

  $\text{1+cos2θ}$ .

According to the Trigonometric formula:

  ${\text{1+cos2θ=2cos}}^{\text{2}}\text{θ}$ .

  ${\text{1+tan}}^{\text{2}}{\text{θ=sec}}^{\text{2}}\text{θ}$ .

  $\Rightarrow {\text{2cos}}^{\text{2}}\text{θ}$ .

  $\Rightarrow \frac{\text{2}}{{\text{sec}}^{\text{2}}\text{θ}}$ .

  $\Rightarrow \frac{\text{2}}{{\text{1+tan}}^{\text{2}}\text{θ}}$ .

$\text{L}\text{.H}\text{.S=}\frac{\text{2}}{{\text{1+tan}}^{\text{2}}\text{θ}}$ ………. $\left(\text{i}\right)$

According to the given:

$\text{R}\text{.H}\text{.S=}\frac{\text{2}}{{\text{1+tan}}^{\text{2}}\text{θ}}$ ……….. $\left(\text{ii}\right)$

By comparing the two equation.

  $\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}$ .

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