Question

Prove that: (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2A + cot^2A

Answer 1

$\textbf{Hey there}$!

$\textbf{The Identities Uesd}$ :

1+tan²A=sec²A

1+cot²A=cosec²A

sin²A+cos²A=1

cosecA=$\dfrac{1}{sinA}$

secA=$\dfrac{1}{cosA}$

$\textbf{Question}$: (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2A + cot^2A

$\textbf{Answer}$ :

=(sinA+cosecA)²+(cosA+secA)²

=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA

=1+cosec²A+sec²A+2+2

=5+(1+cot²A)+(1+tan²A)

=7+tan²A+cot²A


Be Brainly.

Answer 2

Given that,

$(\sin A+cosec A)^2+(\cos A+\sec A)^2=7+\tan^2 A+\cot^2 A$

To prove,

LHS = RHS

Solution,

Taking LHS,

$(\sin A+cosec A)^2+(\cos A+\sec A)^2$

using identity, $(a+b)^2=a^2+b^2+2ab$

$(\sin A+cosec A)^2+(\cos A+\sec A)^2=\sin^2 A+cosec^2+2\sin A cosec A+\cos^2 A+\sec^2 A+2\cos A \sec A$

$\sin A=\dfrac{A}{cosec A}$

So,

$=\sin^2 A+cosec^2+2\sin A \times \dfrac{1}{\sin A}+\cos^2 A+\sec^2 A+2\cos A\times \dfrac{1}{\cos A}\\\\=\sin^2 A+cosec^2+2+\cos^2 A+\sec^2 A+2$

Since, $\sin ^2 A+\cos^2 A=1$

$=cosec^2 A+2+1+\sec^2 A+2$ ....(1)

Now,

$cosec^2 A=\cot^2 A+1\\\\\text{and}\\\\\sec^2 A=1+\tan^2A$

So, equation (1) becomes :

$=\cot^2 A +1+2+1+1+\tan^2 A+2\\\\=7+\tan^2 A +\cot ^2 A\\\\= \text{RHS}$

Therefore, LHS = RHS.