If tan⁡θ=17, then cosec2⁡θ−sec2⁡θcosec2⁡θ+sec2⁡θ is equal tomm+1. the value of m is

tan⁡θ=17⇒cot⁡θ=7Given expression =1+cot2⁡θ−1−tan2⁡θ1+cot2⁡θ+1+tan2⁡θ=cot2⁡θ−tan2⁡θ2+cot2⁡θ+tan2⁡θ=(7)2−1722+(7)2+172=4864=34=mm+1⇒m=3

If tan⁡θ=17, then cosec2⁡θ−sec2⁡θcosec2⁡θ+sec2⁡θ is equal tomm+1. the value of m is

tan⁡θ=17⇒cot⁡θ=7Given expression =1+cot2⁡θ−1−tan2⁡θ1+cot2⁡θ+1+tan2⁡θ=cot2⁡θ−tan2⁡θ2+cot2⁡θ+tan2⁡θ=(7)2−1722+(7)2+172=4864=34=mm+1⇒m=3

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If $\tan θ = \frac{1}{\sqrt{7}}, then (\frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \sec^{2} θ})$ is equal to
$\frac{m}{m + 1}$ . the value of m is

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Detailed Solution

$\tan θ = \frac{1}{\sqrt{7}} \Rightarrow \cot θ = \sqrt{7}$

Given expression $= \frac{1 + \cot^{2} θ - 1 - \tan^{2} θ}{1 + \cot^{2} θ + 1 + \tan^{2} θ}$

$\begin{array}{l} = \frac{\cot^{2} θ - \tan^{2} θ}{2 + \cot^{2} θ + \tan^{2} θ} = \frac{(\sqrt{7})^{2} - {(\frac{1}{\sqrt{7}})}^{2}}{2 + (\sqrt{7})^{2} + {(\frac{1}{\sqrt{7}})}^{2}} \\ = \frac{48}{64} = \frac{3}{4} = \frac{m}{m + 1} \Rightarrow m = 3 \end{array}$