Question

prove that sin0/(sec0+tan0-1) +cos0/(cosec0+cot0-1) = 1​

Answer 1

sinΘ/(1/cosΘ+sinΘ/cosΘ–1) + cosΘ/(1/sinΘ+cosΘ/sinΘ-1)

=sinΘ/ (1+sinΘ-cosΘ)/cosΘ + cosΘ/ (1+cosΘ-sinΘ)/sinΘ

=sinΘcosΘ/(1+sinΘ-cosΘ) + sinΘ cosΘ/(1-sinΘ+cosΘ)

=sinΘcosΘ(1-sinΘ+cosΘ) + sinΘcosΘ(1+sinΘ-cosΘ) / (1+sinΘ-cosΘ)*(1-sinΘ+cosΘ)

=sinΘcosΘ(1-sinΘ+cosΘ + 1+sinΘ-cosΘ) / {(1)^2 – (sinΘ-cosΘ)^2}

=sinΘcosΘ(2) / {1 – (sin^2+ cos^2 – 2sinΘcosΘ)}

=2sinΘcosΘ / {1 – (1 – 2sinΘcosΘ)}

=2sinΘcosΘ / {1-1+2sinΘcosΘ}

=2sinΘcosΘ / 2sinΘcosΘ

=1=RHS. (proved)

Answer 2

To Prove :-

$\sf \dfrac{sin \theta}{sec \theta +tan \theta - 1} + \dfrac{cos \theta }{cosec \theta + cot \theta - 1}= 1$

Solution :-

$\sf L.H.S = \dfrac{sin \theta }{sec \theta + tan \theta -1 } + \dfrac{cos \theta }{cosec \theta + cot \theta -1 }$

$\bullet \bf \ sec \theta = \dfrac{1}{cos \theta }$

$\bullet \bf \ tan \theta = \dfrac{sin \theta }{cos \theta }$

$\bullet \bf \ cosec \theta = \dfrac{1}{sin \theta }$

$\bullet \bf \ cot \theta = \dfrac{cos \theta }{sin \theta }$

  $= \sf \dfrac{sin \theta }{\dfrac{1}{cos \theta }+\dfrac{sin \theta }{cos \theta }-1} + \dfrac{cos \theta }{\dfrac{1}{sin \theta}+\dfrac{cos \theta}{sin \theta }-1} \\\\= \dfrac{sin \theta }{\dfrac{1+sin \theta}{cos \theta}-1 } + \dfrac{cos \theta } {\dfrac{1+cos \theta}{sin \theta} - 1} \\\\= \dfrac{sin \theta}{\dfrac{1+sin \theta - cos \theta }{cos \theta}} + \dfrac{cos \theta }{\dfrac{1+cos \theta - sin \theta}{sin \theta}}$

  $= \sf \left( sin \theta \times \dfrac{cos \theta}{1+sin \theta - cos \theta } \right) + \left( co s \theta \times \dfrac{sin \theta }{1+cos \theta - sin \theta} \right) \\\\= \dfrac{sin \theta cos \theta }{1+sin \theta - cos \theta} + \dfrac{sin \theta cos \theta}{1+cos \theta - sin \theta} \\\\= sin\theta cos \theta \left( \dfrac{1}{1+sin \theta - cos \theta}+\dfrac{1}{1+cos\theta-sin \theta} \right)$

  $= \sf sin\theta cos\theta \left( \dfrac{1+cos\theta - sin \theta +1+sin \theta - cos \theta }{(1+sin \theta - cos \theta )(1+cos \theta - sin \theta )} \right)$                     

  $= \sf sin\theta cos\theta \left( \dfrac{2}{1+cos\theta- sin \theta +sin\theta +sin \theta cos\theta-sin^2 \theta-cos\theta-cos^2 \theta+ sin \theta cos \theta} \right)$

  $= \sf sin \theta cos \theta \left( \dfrac{2}{1+2sin \theta cos \theta -sin^2\theta - cos^2 \theta} \right)$

  $= \sf sin \theta cos \theta \left( \dfrac{2}{1+2sin \theta cos \theta-(sin^2 \theta + cos^2 \theta )} \right)$

$\bullet \bf \ sin^2 \theta + cos ^2 \theta = 1$

      $= \sf sin\theta cos \theta \left( \dfrac{2}{1-1+2sin \theta cos \theta } \right) \\\\= \dfrac{2sin \theta cos \theta }{2 sin \theta cos \theta}\\\\= 1 \\\\= R.H.S$

Hence proved.