Question

Given cot theta =7/8 evaluate
(1+sin theta)(1-sin theta)/(1+cos theta)(1-cos theta)

Answer 1

hi friend,

First we will calculate the value of sinθ and cosθ

Now,tanθ=1/cosθ

tanθ=1/7/8

opp/adj=8/7

BC/AB=8/7

LET BC=8X AND AB=7X

WE find AC using pythagoras therom

hyp²=(height)²+(base)²

AC²=(7x)²+(8x)²

AC²=√113x²

AC=√113x

Now,we need to find sin and cos

here,sin=opp/hyp

=BC/AC=8x/√113x

=8/√113

cos=adj/hyp

AB/AC=7x/√113x

=7/√113

HERE WE HAVE TO EVALUATE

(1+SIN)(1-SIN)/(1+COS)(1-COS)

USING (a+b)(a-b)=a²-b²

=(1²-sin²)/(1²-cos²)

=(1-sin²)/(1-cos²)

PUTTING sin =8/√113 and cos = 7/√113

=[1-(8/√113)²]/[1-(7/√113)²

=(1-8²/√113)/(1-7²/√113)

=(1-64/113)/(1-49/113)

=(113-64/113)/(113-49/113)

=113-64/113-49

=49/64

HENCE,(1+sin)(1-sin)/(1+cos)(1-cos)=49/64

HENCE PROVED✌

Answer 2

hi friend,

First we will calculate the value of sinθ and cosθ

Now,tanθ=1/cosθ

tanθ=1/7/8

opp/adj=8/7

BC/AB=8/7

LET BC=8X AND AB=7X

WE find AC using pythagoras therom

hyp²=(height)²+(base)²

AC²=(7x)²+(8x)²

AC²=√113x²

AC=√113x

Now,we need to find sin and cos

here,sin=opp/hyp

=BC/AC=8x/√113x

=8/√113

cos=adj/hyp

AB/AC=7x/√113x

=7/√113

HERE WE HAVE TO EVALUATE

(1+SIN)(1-SIN)/(1+COS)(1-COS)

USING (a+b)(a-b)=a²-b²

=(1²-sin²)/(1²-cos²)

=(1-sin²)/(1-cos²)

PUTTING sin =8/√113 and cos = 7/√113

=[1-(8/√113)²]/[1-(7/√113)²

=(1-8²/√113)/(1-7²/√113)

=(1-64/113)/(1-49/113)

=(113-64/113)/(113-49/113)

=113-64/113-49

=49/64

HENCE,(1+sin)(1-sin)/(1+cos)(1-cos)=49/64

HENCE PROVED✌