Math, asked by sgowthamraj14, 8 months ago

Prove that( 1+cot² θ )(1 – cos θ)(1 + cos θ) = 1.

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Answered by saniya512
6

Step-by-step explanation:

+ cot2∅)(1 - cos∅)(1 + cos∅) = 1

{use , (a + b)(a - b) = a² - b² }

(1 + cot2∅)( 1 - cos²∅) = 1

[ use, 1 - cos²x = sin²x]

(1 + cot2∅)(sin²∅) = 1

( 1 + tan2∅)(sin²∅) = tan2∅

use, formula,

tan2∅ = 2tan∅/(1 - tan²∅)

( 1 - tan²∅ +2tan∅)sin²∅/(1 - tan²∅) = 2tan∅/( 1 - tan²∅)

(1 -tan²∅+ 2tan∅)sin²∅ = 2tan∅

( 1 - tan²∅+2tan∅)/cosec²∅ = 2tan∅

( 1 - tan²∅ + 2tan∅)/(1 + cot²∅) = 2tan∅

( 1 - tan²∅ + 2tan∅)tan∅/(1 +tan²∅) = 2

tan∅ - tan³∅ +2tan²∅ = 2 + 2 tan²∅

tan³∅ - tan∅ +2 = 0

izvoru47 and 54 more users found this answer helpful

THANKS 31

3.4

(24 votes)

4

It seems you overlooked the question, cause its cot2theta and not cos2theta. Isn't it?

pankaj12je avatar

Abhi its cot^2 theta

abhi178 avatar

but he writes in such a way that anhbody confused

pankaj12je avatar

yah

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pankaj12je

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Hey there !!!

-------------------------------------------------------------------------------------

(1+cot²θ)(1-cosθ)(1+cosθ)=1--------Equation 1

(1+cosθ)(1-cosθ) is of the form (a+b)(a-b)=a²-b²

(1+cosθ)(1-cosθ)=1-cos²θ

But 1-cos²θ=sin²θ

So equation 1 changes to

=(1 + cot²θ )(sin²θ)

=sin²θ+cot²θsin²θ

=sin²θ+(cos²θ*sin²θ/sin²θ)

=sin²θ+cos²θ

=1

LHS=RHS

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Hope this helped you...................

Answered by challapavani1988
2

Answer:

(1+cot^2theta)(1-cos theta)(1+cos theta)= 1

Step-by-step explanation:

  • (1 +cot^2 theta)(1-cos theta)(1+cos theta)=
  • cosec^2theta (1 - cos theta)(1+cos theta)=
  • cosec^2 theta(1-cos^2 theta)=
  • cosec^2 theta(sin^2 theta)=
  • cosec^2 theta(1/cosec ^2theta)=
  • 1
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