Question
Prove the given identities, where the angles involved are acute angles for which the expressions are defined. $(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2A + cot^2A$
✓
Answer
To prove: $(sinA + cosecA)^2 + (cosA + secA)^2 = 7 + tan^2A + cot^2A$
taking L.H.S
Using the formula$ (a+b)^2 = a^2 + b^2 + 2ab to get,$
$= (sin^2A + cosec^2A + 2sinA cosecA) + (cos^2A + sec^2A + 2 cos A sec A)$
Since sin$\theta$ =$\frac {{1 }}{{\ cosec\theta }}$ and $\cos \theta = \frac{1}{{\sec \theta }}$
$=\left(\sin ^{2} A+\csc ^{2} A+2 \sin A \frac{1}{\sin A}\right)+\left(\cos ^{2} A+\sec ^{2} A+2 \cos A \frac{1}{\cos A}\right)$
$= sin^2A + cosec^2A + 2 + cos^2A + sec^2A + 2$
$= (sin^2A + cos^2A) + cosec^2A + sec^2A + 2 + 2$
Using the identities $sin^2A + cos^2A = 1, sec^2A = 1 + tan^2A and cosec^2A = 1 + cot^2A to get$
$= 1+ 1 + tan^2A + 1 + cot^2A + 2 + 2$
$= 1 + 2 + 2 + 2 + tan^2A + cot^2A$
$= 7 + tan^2A + cot^2A$
$= R.H.S.$
Hence proved
taking L.H.S
Using the formula$ (a+b)^2 = a^2 + b^2 + 2ab to get,$
$= (sin^2A + cosec^2A + 2sinA cosecA) + (cos^2A + sec^2A + 2 cos A sec A)$
Since sin$\theta$ =$\frac {{1 }}{{\ cosec\theta }}$ and $\cos \theta = \frac{1}{{\sec \theta }}$
$=\left(\sin ^{2} A+\csc ^{2} A+2 \sin A \frac{1}{\sin A}\right)+\left(\cos ^{2} A+\sec ^{2} A+2 \cos A \frac{1}{\cos A}\right)$
$= sin^2A + cosec^2A + 2 + cos^2A + sec^2A + 2$
$= (sin^2A + cos^2A) + cosec^2A + sec^2A + 2 + 2$
Using the identities $sin^2A + cos^2A = 1, sec^2A = 1 + tan^2A and cosec^2A = 1 + cot^2A to get$
$= 1+ 1 + tan^2A + 1 + cot^2A + 2 + 2$
$= 1 + 2 + 2 + 2 + tan^2A + cot^2A$
$= 7 + tan^2A + cot^2A$
$= R.H.S.$
Hence proved
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