Question

If $\tan \theta +\cot \theta =2$ , find the value of ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta$

Answer 1

Recall that $\tan \theta =\dfrac{1}{\cot \theta }$ .
Square both the sides of the given equation and use the expansion: ${{(a\pm b)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$ .
The value of the square of a real number cannot be negative. i.e. ${{x}^{2}}\ge 0,x\in \mathbb{R}$ .

Complete step-by-step answer:
It is given that $\tan \theta +\cot \theta =2$ .
Squaring both sides, we get:
⇒ ${{(\tan \theta +\cot \theta )}^{2}}={{2}^{2}}$
On expanding by using the distributive property of multiplication, we get:
⇒ ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2\tan \theta \cot \theta =4$
Since $\tan \theta =\dfrac{1}{\cot \theta }$ , we have $\tan \theta \cot \theta =1$ . Substituting this value in the above equation, we get:
⇒ ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2=4$
Subtracting 2 from both the sides, we get:
⇒ ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2$ , which is the required answer.

The answer must be correct, as both ${{\tan }^{2}}\theta ,{{\cot }^{2}}\theta >0$ and $2>0$.

Note: In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
$\sin \theta =\dfrac{P}{H}$ , $\cos \theta =\dfrac{B}{H}$ , $\tan \theta =\dfrac{P}{B}$
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ , $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
$\csc \theta =\dfrac{1}{\sin \theta }$ , $\sec \theta =\dfrac{1}{\cos \theta }$ , $\tan \theta =\dfrac{1}{\cot \theta }$
Some useful algebraic identities:
$(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$
${{(a\pm b)}^{2}}={{a}^{2}}\pm 2ab+{{b}^{2}}$
${{(a\pm b)}^{3}}={{a}^{3}}\pm 3ab(a\pm b)\pm {{b}^{3}}$
$(a\pm b)({{a}^{2}}\mp ab+{{b}^{2}})={{a}^{3}}\pm {{b}^{3}}$