If sec ⁡ θ − tan ⁡ θ sec ⁡ θ + tan ⁡ θ = 1 9 \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{1}{9} sec θ + tan θ sec θ − tan θ ​ =...

Question

If

secθtanθsecθ+tanθ=19\frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{1}{9}
and
θ\theta
lies in the first quadrant, find the value of
sinθtan2θsinθ+tan2θ\frac{\sin \theta - \tan^2 \theta}{\sin \theta + \tan^2 \theta}
.

Options

A.
1327-\frac{13}{27}
B.
1129-\frac{11}{29}
C.
1327\frac{13}{27}
D.
1129\frac{11}{29}
trigonometrysecanttangentfirst quadrantalgebraic manipulation

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