Question

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Answer 1

Question was seen error, so corrected the question a little as follows:

Question:

Prove that tan²θ + cot²θ + 2 = sec²θ · csc²θ

Step-by-step explanation:

⇒ Split 2 in the LHS as 1 + 1 and add both to tan²θ and cot²θ each there. Thus they become sec²θ and csc²θ each.

⇒ Take the reciprocals of the both (sec²θ as 1/cos²θ and csc²θ as 1/sin²θ) and add them.

⇒ The numerator sin²θ + cos²θ becomes 1 and the fraction will be taken its reciprocal to get the RHS.

⇒ Hence proved!

Method:

$\Rightarrow\ LHS \\ \\ \Rightarrow\ \tan^2\theta+\cot^2\theta+2 \\ \\ \Rightarrow\ \tan^2\theta+\cot^2\theta+1+1 \\ \\ \Rightarrow\ \tan^2\theta+1+\cot^2\theta+1 \\ \\\Rightarrow\ \sec^2\theta+\csc^2\theta \\ \\ \Rightarrow\ \frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta} \\ \\ \Rightarrow\ \frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta \cdot \sin^2\theta} \\ \\ \Rightarrow\ \frac{1}{\cos^2\theta \cdot \sin^2\theta} \\ \\ \Rightarrow\ \sec^2\theta \cdot \csc^2\theta \\ \\ \Rightarrow\ RHS$