Question

prove that -
sec^2{tan^1(2)}+cosec^2{cot^1(3)=15

Answer 1

sec2 { sec -1 root 5 } + cosec2 { cosec -1 root 10}
{ using pythogeros }
{root 5 }2 +{ root 10} 2 = 15
May this help you

Answer 2

Answer:

It is proved that $\bold{\begin{equation}\sec ^{2}\left\{\tan ^{1}(2)\right\}+\csc ^{2}\left\{\cot ^{1}(3)\right\}=15\end{equation}}$

Step-by-step explanation:

According to Trigonometric Identities we get

$\begin{array}{l}{\sec ^{2}(x)=\tan ^{2}(x)+1} \\ {\csc ^{2}(x)=\cot ^{2}(x)+1}\end{array}$

So, we take the value of $\tan ^{-1}(2)=a \text { and } \cot ^{-1}(3)=b$

Taking LHS separately and substituting,

Putting  $\tan ^{-1}(2)=a \text { and } \cot ^{-1}(3)=b$ in $\begin{equation}\sec ^{2}\left\{\tan ^{-1}(2)\right\}+\csc^{2}\left\{\cot ^{-1}(3)\right\}\end{equation}$

We get

$\begin{equation}\begin{array}{l}{\sec ^{2}\{a\}+\csc ^{2}\{b\}=15} \\ {\tan ^{2}(a)+1+\cot ^{2}(b)+1=15}\end{array}\end{equation}$

Now, substituting the values we get,

$\begin{equation}\tan ^{2}\left(\tan ^{-1}(2)\right)+1+\cot ^{2}\left(\cot ^{-1}(3)\right)+1=15\end{equation}$

$\begin{equation}\begin{array}{l}{\left(\tan \left(\tan ^{-1}(2)\right)^{2}\right)+1+\left(\cot \left(\cot ^{-1}(3)\right)^{2}\right)+1=15} \\ {(2)^{2}+1+(3)^{2}+1=15}\end{array}\end{equation}$

4 + 1 + 9 + 1 = 15

Hence it is proved that LHS is equal to RHS (15 = 15).