Question

Answer Please !!!!!!​

Answer 1

To Prove :-

$\sf ( cosec \theta - sin \theta )(sec \theta - cos \theta )(tan \theta + cot \theta ) = 1$

Solution :-

$\sf L.H.S = ( cosec \theta - sin \theta )( sec \theta - cos \theta )( tan \theta + cot \theta )$

$\bullet \bf \ cosec \theta = \dfrac{1}{sin \theta } \\\\\bullet \ sec \theta = \dfrac{1}{cos \theta } \\\\\bullet \ tan \theta = \dfrac{sin \theta }{cos \theta } \\\\\bullet \ cot \theta = \dfrac{cos \theta }{sin \theta }$

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

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

      $= \sf \dfrac{cos^2 \theta }{sin \theta } \times \dfrac{sin^2 \theta }{cos \theta } \times \dfrac{1}{sin \theta cos \theta }\\\\= \dfrac{cos \theta sin \theta }{sin \theta cos \theta } \\\\= 1 \\\\= R.H.S$

Hence proved.