1.

If tan θ = \(\frac{1}{{\sqrt 7 }}\) and 0° < θ < \(\frac{\pi }{2},\) then the value of \(\frac{{co{t^2}\theta - {{\sec }^2}\theta }}{{se{c^2}\theta + {{\cot }^2}\theta }}\) is1). \(\frac{{36}}{{51}}\)2). \(\frac{{33}}{{54}}\)3). \(\frac{{41}}{{57}}\)4). \(\frac{{46}}{{61}}\)

Answer»

Given,

tan θ = 1/√7

? tan X = Opposite side/Adjacent side

Opposite side = 1 unit

Adjacent side = √7 units

By Pythagorean Theorem,

(Hypotenuse)2 = (Side)2 + (Side)2

⇒ Hypotenuse2 = 12 + (√7)2 = 8

⇒ Hypotenuse = √8 = 2√2

$(\BEGIN{array}{L} \frac{{co{t^2}\theta - {{\sec }^2}\theta }}{{SE{c^2}\theta + {{\cot }^2}\theta }}\; = \;\frac{{7 - \frac{8}{7}}}{{\frac{8}{7} + 7}}\; = \;\frac{{41}}{{57}}\\ \therefore \frac{{co{t^2}\theta - {{\sec }^2}\theta }}{{se{c^2}\theta+{{\cot }^2}\theta}}=\frac{{41}}{{57}}\END{array})$



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