Math, asked by Anonymous, 1 year ago

Prove that (cot θ + cosec θ - 1 ) / (cot θ - cosec θ + 1) = (1 + cos θ) / sinθ

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Answered by Anonymous
24

HEY MATE YOUR ANSWER IS HERE...

SORRY FOR DELAYING...

ANSWER IS 100% CORRECT BUT AT MANY TIMES I HAD RATIONALISED THE EQUATION...

AND TOO MANY IDENTITIES ARE USED.....THANKS

KEEP SMILING ☺️☺️☺️✌️

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Answered by Anonymous
19

Solution :-

 \rm  \dfrac{cot \theta + cosec \theta - 1}{cot \theta - cosec \theta + 1}  =  \dfrac{1 + cos \theta}{sin \theta}

Consider LHS

 \rm  =  \dfrac{cot \theta + cosec \theta - 1}{cot \theta - cosec \theta + 1}

Writing cot Φ in terms of sin Φ and cosΦ

 \rm  =  \dfrac{ \dfrac{cos \theta}{sin \theta}  +  \dfrac{1}{sin \theta} - 1}{ \dfrac{cos \theta}{sin \theta}  -  \dfrac{1}{sin \theta}  + 1}

Taking LCM

 \rm  =  \dfrac{ \dfrac{cos \theta + 1 - sin \theta}{sin \theta} }{ \dfrac{cos \theta - 1 + sin \theta}{sin \theta} }

 \rm  =  \dfrac{cos \theta + 1 - sin \theta}{cos \theta - 1 + sin \theta}

Rearranging the terms

 \rm  =  \dfrac{cos \theta - sin \theta + 1}{cos \theta + sin \theta - 1}

Rationalising the denominator

 \rm  =  \dfrac{cos \theta -sin \theta   +  1}{cos \theta + sin \theta - 1} \times  \dfrac{cos \theta + sin \theta + 1}{cos \theta + sin \theta + 1}

 \rm  =  \dfrac{cos \theta   +  1 - sin \theta}{cos \theta + sin \theta - 1} \times  \dfrac{cos \theta  + 1+ sin \theta}{cos \theta + sin \theta + 1}

 \rm  =  \dfrac{(cos \theta   +  1 ) ^{2} - sin^{2}  \theta}{(cos \theta + sin \theta)^{2}  - 1}

 \rm  =  \dfrac{cos^{2}  \theta  + 2cos \theta  +  cos^{2}  \theta}{1+ 2cos \theta sin \theta - 1}

 \rm  =  \dfrac{2cos^{2}  \theta  + 2cos \theta  }{2cos \theta sin \theta }

 \rm  =  \dfrac{2cos \theta(1 + cos \theta)  }{2cos \theta sin \theta }

 \rm  =  \dfrac{1 + cos \theta}{ sin \theta }

= RHS

LHS = RHS

QED

Identities used to prove

  • cos² Φ + sin²Φ = 1

  • cos² Φ = 1 - sin² Φ

  • x² - y² = (x + y)(x - y)

  • (x + y)² = x² + y² + 2xy
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