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$I f 0 < θ < \frac{π}{2}, \tan θ = \frac{\cos 29 ° + \sin 29 °}{\cos 29 ° - \sin 29 °}, t h e n θ =$

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Correct option is B

We have tanθ=cos29∘+sin29∘cos29∘-sin29∘ =1+tan29∘1-tan29∘ =tan45°+29° =tan74°

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If 0<θ<π2, tanθ=cos29°+sin29°cos29°−sin29°, then θ=

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$I f 0 < θ < \frac{π}{2}, \tan θ = \frac{\cos 29 ° + \sin 29 °}{\cos 29 ° - \sin 29 °}, t h e n θ =$