# If tanA+sinA=p and tanA-sinA=q. Then proove that p^2-q^2=4sqrt (pq) ?

Dec 28, 2017

It is proved below:

#### Explanation:

Here,
$\tan A + \sin A = p$

$\tan A - \sin A = q$

$L . H . S = {p}^{2} - {q}^{2}$

$= {\left(\tan A + \sin A\right)}^{2} - {\left(\tan A - \sin A\right)}^{2}$

$= \left(\tan A + \cancel{\sin} A + \tan A - \cancel{\sin} A\right) \left(\cancel{\tan} A + \sin A - \cancel{\tan} A + \sin A\right)$

$= \left(2 \tan A\right) \cdot \left(2 \sin A\right)$

$= 4 \tan A \sin A$

$= 4 \sqrt{{\tan}^{2} A {\sin}^{2} A}$

$= 4 \sqrt{\left({\sec}^{2} A - 1\right) \sin A}$$\textcolor{b l u e}{\left[\left[A s , {\sec}^{2} \theta - {\tan}^{2} \theta = 1\right]\right]}$

$= 4 \sqrt{{\sec}^{2} A . {\sin}^{2} A - {\sin}^{2} A}$

$= 4 \sqrt{{\sin}^{2} \frac{A}{{\cos}^{2} A} - {\sin}^{2} A}$

$= 4 \sqrt{{\tan}^{2} A - {\sin}^{2} A}$

$= 4 \sqrt{\left(\tan A + \sin A\right) \left(\tan A - \sin A\right)}$

$= 4 \sqrt{p q}$$\textcolor{g r e e n}{\left[A s . \tan A + \sin A = p \mathmr{and} \tan A - \sin A = q\right]}$

$= R . H . S$

Dec 30, 2017

Kindly refer to the Explanation.

#### Explanation:

$\tan A + \sin A = p \ldots \ldots \left(1\right) , \mathmr{and} , \tan A - \sin A = q \ldots \ldots \left(2\right)$.

$\therefore \left(1\right) + \left(2\right) \Rightarrow 2 \tan A = p + q , \mathmr{and} , \tan A = \frac{p + q}{2}$.

$\text{Similarly, } \sin A = \frac{p - q}{2}$.

$\therefore \cot A = \frac{2}{p + q} , \mathmr{and} \csc A = \frac{2}{p - q}$.

But, ${\csc}^{2} A - {\cot}^{2} A = 1$,

$\Rightarrow \frac{4}{p - q} ^ 2 - \frac{4}{p + q} ^ 2 = 1 , \mathmr{and} ,$.

$\frac{4 \left\{{\left(p + q\right)}^{2} - {\left(p - q\right)}^{2}\right\}}{{\left(p + q\right)}^{2} {\left(p - q\right)}^{2}} = 1$.

$\Rightarrow \frac{4 \cdot 4 p q}{{p}^{2} - {q}^{2}} ^ 2 = 1 , i . e . ,$

${\left({p}^{2} - {q}^{2}\right)}^{2} = 16 p q$,

$\Rightarrow {p}^{2} - {q}^{2} = 4 \sqrt{p q} ,$ as desired!

Enjoy Maths.!

Dec 30, 2017

$R H S = 4 \sqrt{p q}$

$= 4 \sqrt{\left(\tan A + \sin A\right) \left(\tan A - \sin A\right)}$

$= 4 \sqrt{{\tan}^{2} A - {\sin}^{2} A}$

$= 4 \sqrt{{\sin}^{2} \frac{A}{\cos} ^ 2 A - {\sin}^{2} A}$

$= 4 \sqrt{{\sin}^{2} A \times {\sec}^{2} A - {\sin}^{2} A}$

=4sqrt(sin^2A(sec^2A-1)

$= 4 \sqrt{{\sin}^{2} A {\tan}^{2} A}$

$= 4 \tan A \sin A$

$= {\left(\tan A + \sin A\right)}^{2} - {\left(\tan A - \sin A\right)}^{2}$

$= {p}^{2} - {q}^{2} = R H S$