Question

If tan (π/4+theta) + tan (π/4 - theta) = a, then
tan square (π/4 + theta) + tan square (π/4-theta) is ????
​

Answer 1

AnswEr :

The value of above expression is a² - 2

Given that,

$\sf \: \tan( \dfrac{\pi}{4} + \alpha ) + \tan( \dfrac{\pi}{4} - \alpha ) = a - - -- - - - (1)$

To finD

$\tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha )$

Squaring equation (1) on both sides,

$\sf \: \tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha ) + 2 \tan( \dfrac{\pi}{4} + \alpha) \tan( \dfrac{\pi}{4} - \alpha ) = {a}^{2} \\ \\ \longrightarrow \: \sf \: \tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha ) + 2 \bigg( \dfrac{ \tan( \dfrac{\pi}{4} ) + \tan( \alpha ) }{1 - \tan( \dfrac{\pi}{4} ) \tan( \alpha ) } \bigg) \bigg( \dfrac{ \tan( \dfrac{\pi}{4} ) - \tan( \alpha ) }{1 + \tan( \dfrac{\pi}{4} ) \tan( \alpha ) } \bigg) = {a}^{2} \\ \\ \longrightarrow \: \sf \: \tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha ) + 2 \times \dfrac{ 1 + \tan( \alpha ) }{1 - \tan( \alpha ) } \times \dfrac{1 - \tan( \alpha ) }{1 + \tan( \alpha ) } = {a}^{2} \\ \\ \longrightarrow \: \sf \: \tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha ) + 2 = {a}^{2} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \tan {}^{2} ( \dfrac{\pi}{4} + \alpha ) + \tan {}^{2} ( \dfrac{\pi}{4} - \alpha ) = {a}^{2} - 2}}$

Identities Used :

• $\sf tan(a + b) = \dfrac{tan(a) + tan (b)}{1 - tan(a)tan(b)}$

• $\sf tan(a - b) = \dfrac{tan(a) - tan (b)}{1 + tan(a)tan(b)}$

• tan(π/4) = 1

Answer 2

Answer:

$\underline{\bf Given:}\:\rm\tan\left(\dfrac{\pi}{4} +\theta\right) +\tan\left(\dfrac{\pi}{4} - \theta\right) = a\\\\\underline{\bf To\:Find:}\:\tan^2\left(\dfrac{\pi}{4} +\theta\right) +\tan^2\left(\dfrac{\pi}{4} - \theta\right) = ?$

$\rm Let, \quad\tan\left(\dfrac{\pi}{4} +\theta\right)=x \\\\\rm Then,\quad\tan\left(\dfrac{\pi}{4} - \theta\right) = \dfrac{1}{x}$

• As we can see that both are Opposite of each other.

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\sf \tan\left(\dfrac{\pi}{4} +\theta\right) +\tan\left(\dfrac{\pi}{4} - \theta\right) = a\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Taking value in x.}}\\\\:\implies\sf x+\dfrac{1}{x}=a\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Squaring Both Sides.}}\\\\:\implies\sf \left(x+\dfrac{1}{x}\right)^2=(a)^2\\\\\\:\implies\sf x^2+\left(\dfrac{1}{x}\right)^2+\left(2 \times x \times \dfrac{1}{x}\right)=a^2\\\\\\:\implies\sf x^2 + \dfrac{1}{x^2} + 2 = a^2\\\\\\:\implies\sf x^2 + \dfrac{1}{x^2} =a^2 - 2\\\\\\:\implies\underline{\boxed{\sf \tan^2\left(\dfrac{\pi}{4} +\theta\right) +\tan^2\left(\dfrac{\pi}{4} - \theta\right) = a^2 - 2}}$