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Question: If \(x = \sec\varphi - \tan\varphi,y = \text{cosec}\varphi + \cot\varphi,\) then...

If x=sec⁡φ−tan⁡φ,y=cosecφ+cot⁡φ,x = \sec\varphi - \tan\varphi,y = \text{cosec}\varphi + \cot\varphi, then

A

x=y+1y−1x = \frac{y + 1}{y - 1}

B

x=y−1y+1x = \frac{y - 1}{y + 1}

C

y=1−x1+xy = \frac{1 - x}{1 + x}

D

None of these

Answer

x=y−1y+1x = \frac{y - 1}{y + 1}

Explanation

Solution

We have xy=(sec⁡φ−tan⁡φ)(cosecφ+cot⁡φ)xy = (\sec\varphi - \tan\varphi)(\text{cosec}\varphi + \cot{}\varphi)

=1−sin⁡φcos⁡φ.1+cos⁡φsin⁡φ= \frac{1 - \sin\varphi}{\cos\varphi}.\frac{1 + \cos\varphi}{\sin\varphi}

⇒xy+1=1−sin⁡φ+cos⁡φ−sin⁡φcos⁡φ+sin⁡φcos⁡φcos⁡φsin⁡φ\Rightarrow xy + 1 = \frac{1 - \sin\varphi + \cos\varphi - \sin\varphi\cos\varphi + \sin\varphi\cos\varphi}{\cos\varphi\sin\varphi}

=1−sin⁡φ+cos⁡φcos⁡φsin⁡φ= \frac{1 - \sin\varphi + \cos\varphi}{\cos\varphi\sin\varphi} …..(i)

x−y=(sec⁡φ−tan⁡φ)−(cos⁡⥂⥂ecφ+cot⁡φ)x - y = (\sec\varphi - \tan\varphi) - (\cos ⥂ ⥂ ec\varphi + \cot\varphi)

=1−sin⁡φcos⁡φ−1+cos⁡φsin⁡φ=sin⁡φ−sin⁡2φ−cos⁡φ−cos⁡2φcos⁡φsin⁡φ= \frac{1 - \sin\varphi}{\cos\varphi} - \frac{1 + \cos\varphi}{\sin\varphi} = \frac{\sin\varphi - \sin^{2}\varphi - \cos\varphi - \cos^{2}\varphi}{\cos\varphi\sin\varphi}

=sin⁡φ−cos⁡φ−1cos⁡φsin⁡φ= \frac{\sin\varphi - \cos\varphi - 1}{\cos\varphi\sin\varphi} …..(ii)

Adding (i) and (ii) we get, xy+1+(x−y)=0xy + 1 + (x - y) = 0

⇒x=y−1y+1\Rightarrow x = \frac{y - 1}{y + 1}.