Question

sec square tan inverse 2 + cosec square cot inverse 3= 15

Answer 1

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Answer 2

Answer and Explanation:

To prove : $\sec^2(\tan^{-1}2)+\csc^2(\cot^{-1}3)=15$

Proof :

Taking left hand side,

$LHS=\sec^2(\tan^{-1}2)+\csc^2(\cot^{-1}3)$

Applying trigonometry properties,

$\sec^2 \theta=1+\tan^2\theta$ and $\csc^2 \theta=1+\cot^2\theta$

$LHS=1+\tan^2 (\tan^{-1}2)+1+\cot^2(\cot^{-1}3)$

We know, $\tan (\tan^{-1}x)=x,\ \cot (\cot^{-1}x)=x,$

$LHS=2+(2)^2+1+(3)^2$

$LHS=2+4+1+9$

$LHS=15$

$LHS=RHS$

Therefore, $\sec^2(\tan^{-1}2)+\csc^2(\cot^{-1}3)=15$