Question

(sin a+cosec a)^2+(cos a+sec a)^2=5+ sec^2 a.cosec^2 a

Answer 1

(sin^2A+cosec^2A+2.sinA.cosecA)+(cos^2A+... 
=(sin^2A+cos^2A)+(2.sinA.cosecA)+(2.co... 
=1+2+2+(cosec^2A+sec^2A) 
=5+((sin^2A+cos^2A)/(cos^2A.sin^2A)) 
=5+(sec^2A.cosec^2A) 
=5+(1+tan^2A)(1+cot^2A) 
=5+1+cot^2A+tan^2A+tan^2A.cot^2A 
=5+1+1+cot^2A+tan^2A 
=7+tan^2A+cot^2A 
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Answer 2

$(sin \: a + cosec \: a) {}^{2} + (cos \: a + sec \: a) {}^{2}$


Opening the brackets, using identity (a+b) ^2 = a^2 +b^2 +2ab


$\implies {sin}^{2} a \: + {cosec}^{2} a + 2 \: sin \: a \: \times cosec \: a \\ + {cos}^{2} a \: + sec {}^{2} a \: + 2 \: cos \: a \times \sec \: a \:$




$\implies \: \sin {}^{2}a + \cos {}^{2} a + cosec {}^{2} a + \\ \sec {}^{2} a + 2 \times \cancel{ \sin \: a }\times \frac{1}{ \cancel{ \sin \: a}} + \\ 2 \times \cancel{ \cos \: a }\times \frac{1}{ \cancel{ \cos \: a} }$


As sin^2 a+cos^2 a=1

$\implies \: 1 + \cosec {}^{2}a + \sec {}^{2} a + 2 + 2$


$\implies \: 5 + 1 + { \tan }^{2} a + 1 + { \cot}^{2} a$


As cosec^2 A = 1+tan^2 A

and sec^2 A = 1+ cot^2 A

$\implies \: 7 + { \tan}^{2} a + { \cot} ^{2} a$