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Mathematics

If tan θ = (17)\Big(\dfrac{1}{\sqrt{7}}\Big), then the value of (cosec2θ+sec2θcosec2θsec2θ)\Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) is:

  1. (34)\Big(\dfrac{3}{4}\Big)

  2. (43)\Big(\dfrac{4}{3}\Big)

  3. (37)\Big(\dfrac{3}{7}\Big)

  4. (47)\Big(\dfrac{4}{7}\Big)

Trigonometric Identities

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Answer

Given,

tan θ = (17)\Big(\dfrac{1}{\sqrt{7}}\Big)

tan2 θ = (17)\Big(\dfrac{1}{7}\Big)

We know that,

sec2θ=1+tan2θsec2θ=1+17sec2θ=87.cosec2θ=1+cot2θcosec2θ=1+7cosec2θ=8.\sec^2 \theta = 1 + \tan^2 \theta \\[1em] \sec^2 \theta = 1 + \dfrac{1}{7} \\[1em] \sec^2 \theta = \dfrac{8}{7}. \\[1em] \cosec^2 \theta = 1 + \cot^2 \theta \\[1em] \cosec^2 \theta = 1 + 7 \\[1em] \cosec^2 \theta = 8.

Given expression,

(cosec2θ+sec2θcosec2θsec2θ)8+8788756+8756876448=43.\Rightarrow \Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) \\[1em] \Rightarrow \dfrac{8 + \dfrac{8}{7}}{8 - \dfrac{8}{7}} \\[1em] \Rightarrow \dfrac{\dfrac{56 + 8}{7}}{\dfrac{56 - 8}{7}} \\[1em] \Rightarrow \dfrac{64}{48} = \dfrac{4}{3}.

Hence, option 2 is the correct option.

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