Question

TanB= n sin a.cos a/1-nSin²a


prove that tan (a - b) = (1 - n) tan a.
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Answer 1

Step-by-step explanation:

tanb= nsina cosa/1 - n sin^2a

tan( a- b)= tana - tanb/1+ tana tanb

={ tana - (nsiacosa/1-nsina^2a ]/1+ tana ( nsina cosa/1- nsin^2a)

= [tana*(1- nsin^2a) - nsina cosa]/ {1- nsin^2a +tana(nsina cosa }

=tana - nsin^2a *tana - nsina cosa]/1- nsin^2a + n sin^2a

= tana[ 1- nsin^2 - ncos^2a]/1

= tana[ 1- n(sin^2a + cos^2a) ]

= tana ( 1- n) proved

Answer 2

★Given:-

• Tan B= nsin A.cos A/1-nSin²A

★To prove:-

• tan (A - B) = (1 - n) tan A

★Proof:-

We know,

$\displaystyle \leadsto \underline{\boxed{\sf tan(A-B) = \frac{tanA-tanB}{1+tanAtanB}}}$

Putting the formula,

✦Tanθ = Sinθ/cosθ

$\\ \displaystyle \implies \sf \frac{\bigg( \cfrac{sinA}{cosA} - \cfrac{nsinAcosA}{1-nsin^2A} \bigg) }{1+\bigg( \cfrac{sinA}{cosA} \times \cfrac{nsinAcosA}{1-nsin^2A} \bigg)}$

$\displaystyle \implies \sf \frac{\cfrac{sinA-nsin^3A - nsinAcos^2A}{cosA(1-nsin^2A)} }{ 1+\bigg(\cfrac{nsin^2AcosA}{cosA(1-nsin^2A)}\bigg) }$

$\displaystyle \implies \sf \frac{sinA-nsin^3A-nsinAcos^2A}{cosA(1-nsin^2A)+nsin^2AcosA}$

Putting the formula,

✦Cos²A = 1-sin²A

$\\ \displaystyle \implies \sf \frac{sinA-nsin^3A-nsinA(1-sin^2A)}{cosA-nsin^2AcosA+nsin^2AcosA}$

$\displaystyle \implies \sf \frac{sinA\cancel{-nsin^3A}-nsinA \cancel{+nsin^3A}}{cosA\cancel{-nsin^2AcosA}\cancel{+nsin^2AcosA}}$

$\displaystyle \implies \sf \frac{sinA-nsinA}{cosA}$

$\displaystyle \implies \sf \frac{sinA(1-n)}{cosA}$

$\displaystyle \implies \underline{\boxed{\sf tanA(1-n).}}$

Hence proved!

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