Question

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

Answer 1

Hint: 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}} $