Question

prove that (sin2A+cos2A) + (cos2A+sec2A) = 7+tan2A+cot2A

Answer 1

Step-by-step explanation:

bic main + nahi hoga. please check the question

Answer 2

Answer with Step-by-step explanation:

LHS

$(sinA+cosecA)^2+(cosA+secA)^2$

$sin^2A+cosec^2A+2sinAcosecA+cos^2A+sec^2A+2cosAsecA$

Using identity :$(a+b)^2=a^2+b^2+2ab$

$sin^2A+1+cot^2A+2sinA\times \frac{1}{sinA}+cos^2A+1+tan^2A+2cosA\times \frac{1}{cosA}$

Using formula:$1+tan^2A=sec^2A,1+cosec^2A=cot^2A,cosecA=\frac{1}{sinA},secA=\frac{1}{cosA}$

$1+1+1+2+2+tan^2A+cot^2A$

Using $sin^2A+cos^2A=1$

$7+tan^2A+cot^2A$

LHS=RHS

Hence, proved.

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