Question

prove that (sinA+cosecA)^2+(cosA+secA)^2=7+tan^2A+cot^2A

Answer 1

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Answer 2

$\bf{Answer\colon}$

$Trigonometric\:identies\:used\colon\\sin^2A+cos^2A=1\\1+tan^2A=sec^2A\\1+cot^2A=cosec^2A\\\frac{1}{sinA}=cosecA\\\frac{1}{cosA}=cosecA$

$LHS\\\\={(sin\:A+cosec\:A)}^{2}\\+{(cosA+secA)}^{2}\\\\=sin^2A+cosec^2A\\+2\:sinA\:cosecA+cos^2A\\+sec^2A+2\:cosA\:secA\\\\=(sin^2A+cos^2A)\\+(1+cot^2A)+2\:sinA\times\frac{1}{sinA}\\+(1+tan^2A)+2\:cosA\times\frac{1}{cosA}\\\\=1+(1+cot^2A)+2\\+(1+tan^2A)+2\\\\=7+tan^2A+cot^2A\\\\=RHS.\\\\Hence\:Proved$