If -178- then the value of sin-1- 2-cos-1-tan 2- is

The correct option is C 0Given expression sinθ|cosec θ|+cosθ|θ| =sinθ cosec θ+cosθ( θ) (∵θ∈ Q_2∴cosec θ gt;0 , θ lt;0) =1 1=0 ...

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If θ=178∘, th...

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If $θ = 178 °$ , then the value of $sin θ \sqrt{1 + {cot}^{2} θ} + cos θ \sqrt{1 + {tan}^{2} θ}$ is

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\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

{sec}^{2} θ = 1 + {tan}^{2} θ

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sin θ - \tan θ}

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}

cos θ = \frac{1}{2}

\frac{2 sec θ}{1 + {tan}^{2} θ}

{cot}^{2} θ = \frac{7}{8}

0 < θ < 90^{o}

\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}

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