Question

If \((\tan\theta+\cot\theta)=6\), then the value of \(\tan^{2}\theta+\cot^{2}\theta\) is

• A. \(25\)
• B. \(27\)
• C. \(24\)
• D. \(34\)

Hint:

Whenever \(x+\frac{1}{x}\) is known, squaring gives \(x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2\).

Answer 1

The Correct Option is D

Step 1: Use identity.
Let \(x=\tan\theta\). Then \(x+\dfrac{1}{x}=6\).
Step 2: Square both sides.
\(\left(x+\dfrac{1}{x}\right)^{2}=x^{2}+2+\dfrac{1}{x^{2}}=36\Rightarrow x^{2}+\dfrac{1}{x^{2}}=34.\)
Thus \(\tan^{2}\theta+\cot^{2}\theta=34\).