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The value of secθ(1+sinθcosθ+cosθ1+sinθ)2tan2θsec hetaleft(frac{1+sin heta}{cos heta}+frac{cos heta}{1+sin heta} ight)-2 an ^{2} heta is

a

0

b

1

c

4

d

2

Answer : Option D
Explanation :
Expression, =secθ(1+sinθcosθ+cosθ1+sinθ)2tan2θ=sec hetaleft(frac{1+sin heta}{cos heta}+frac{cos heta}{1+sin heta} ight)-2 an ^{2} heta =1+sin2θ+2sinθ+cos2θcos2θ(1+sinθ)2tan2θ=frac{1+sin ^{2} heta+2 sin heta+cos ^{2} heta}{cos ^{2} heta(1+sin heta)}-2 an ^{2} heta =2+2sinθcos2θ(1+sinθ)2tan2θ=frac{2+2 sin heta}{cos ^{2} heta(1+sin heta)}-2 an ^{2} heta =2cos2θ2tan2θ=2sec2θ2tan2θ=frac{2}{cos ^{2} heta}-2 an ^{2} heta=2 sec ^{2} heta-2 an ^{2} heta =2(sec2θtan2θ)=2=2left(sec ^{2} heta- an ^{2} heta ight)=2

विकल्प:

माना = 30°

Expression =secθ(1+sinθcosθ+cosθ1+sinθ)2tan2θ=sec hetaleft(frac{1+sin heta}{cos heta}+frac{cos heta}{1+sin heta} ight)-2 an ^{2} heta

=sec30(1+sin30cos30+cos301+sin30)2tan230=sec 30^{circ}left(frac{1+sin 30^{circ}}{cos 30^{circ}}+frac{cos 30^{circ}}{1+sin 30^{circ}} ight)-2 an ^{2} 30^{circ}

=23(1+1232+321+12)2×(13)2=frac{2}{sqrt{3}}left(frac{1+frac{1}{2}}{frac{sqrt{3}}{2}}+frac{frac{sqrt{3}}{2}}{1+frac{1}{2}} ight)-2 imesleft(frac{1}{sqrt{3}} ight)^{2}

=23(33+33)23=23×4323=frac{2}{sqrt{3}}left(frac{3}{sqrt{3}}+frac{sqrt{3}}{3} ight)-frac{2}{3}=frac{2}{sqrt{3}} imes frac{4}{sqrt{3}}-frac{2}{3}

=63=2=frac{6}{3}=2

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