Question

`sec {}^(2) a \: . \: cosec {}^(2) a = {tan}^(2)a + {cot}^(2) a + 2 \\`

Please help!!!!!!!!!!!!​

Answer 1

See below ~

Explanation:

Trigonometric Identies (needed)

• sec²a = 1 + tan²a
• cosec²a = 1 + cot²a
• cot²a = 1/tan²a

Solving

Expanding the LHS

• sec²a * cosec²a
• (1 + tan²a) * (1 + cot²a)
• 1 + tan²a + cot²a + (tan²a * cot²a)
• 1 + tan²a + cot²a + 1
• tan²a + cot²a + 2
• ⇒ RHS

∴ Hence, we have proved the left hand side of the equation is equal to the right hand side.

Answer 2

Trigonometric identities

`\sec^2(\alpha)=1+\tan^2(\alpha)`

`\csc^2(\alpha)=1+\cot^2(\alpha)`

`\cot^2(\alpha)=(1)/(\tan^2(\alpha))`

Solution

`\begin{aligned}\sec^2(\alpha) \cdot \csc^2(\alpha) & = (1+\tan^2(\alpha))(1+\cot^2(\alpha))\\\\ & =1+\cot^2(\alpha)+\tan^2(\alpha)+tan^2(\alpha)\cot^2(\alpha)\\\\ & = 1+\cot^2(\alpha)+\tan^2(\alpha)+\tan^2(\alpha) \cdot (1)/(\tan^2(\alpha))\\\\ & = 1+\cot^2(\alpha)+\tan^2(\alpha)+ (\tan^2(\alpha))/(\tan^2(\alpha))\\\\& = 1+\cot^2(\alpha)+\tan^2(\alpha)+1\\\\& = \tan^2(\alpha)+\cot^2(\alpha)+2\end{aligned}`