Math, asked by yamar7020566, 10 months ago

tan[π/4+A]+tan[π/4-A] divide by tan[π/4+A]-tan[π/4-A] =cosec2A​

Answers

Answered by spiderman2019
0

Answer:

Step-by-step explanation:

tan[π/4+A]+tan[π/4-A]/tan[π/4+A]-tan[π/4-A]

//Solve numerator and denominator separately

tan[π/4+A]+tan[π/4-A]

//Tan (A+B) = TanA+ TanB / 1- TanATanB;

//Tan (A-B) = TanA - TanB / 1 + TanATanB

=> Tan(π/4) + TanA/ 1 - Tan{π/4}TanA + Tan(π/4) -  TanA/ 1 + Tan{π/4}TanA

=> 1 + TanA/ 1 - TanA + 1 - TanA/1+TanA

= 2(1+Tan²A) / 1 - Tan²A  

//Now for denominator

tan[π/4+A]-tan[π/4-A]

=>  Tan(π/4) + TanA/ 1 - Tan{π/4}TanA -  Tan(π/4) -  TanA/ 1 + Tan{π/4}TanA

=> 1 + TanA/ 1 - TanA -  1 - TanA/1+TanA

= 2(2TanA) / 1 - Tan²A

//Substitute back the values in main equation

2(1+Tan²A) / 1 - Tan²A /  2(2TanA) / 1 - Tan²A

= 2Sec²A /4TanA

=1/2sincosa

= 1/Sin2A

= Coesc2a

= R.H.S

Hence Proved

Answered by Anonymous
0

 \tan = ( \frac{ \pi}{4}  + a)  \times  \tan( \frac{3 \pi + A}{4} )

using \: tangent \: sum \: of \: angle \: formula \ratio -

 tan(A+B) =  \frac{ \tan A+ \tan B}{1 -  \tan A \times  \tan B }

 in \: conjuction \: with \ratio

\tan( \frac{ \tan \pi }{4} )  = 1 \: and \: \tan( \frac{ \tan 3\pi }{4} ) =  - 1

now

\tan(  \frac{\pi}{4}   + A) \times  \tan( \frac{3 \pi }{4}  + A)

=  \frac{ \tan (\frac{ \pi }{4} ) +  \tan  A}{1 -  \tan( \frac{ \pi }{4}) \times  \tan A  }

=  > \frac{ \tan (\frac{3 \pi }{4} ) +  \tan  A}{1 -  \tan( \frac{ 3\pi }{4}) \times  \tan A  }

=   \frac{1 +  \tan A }{1 -  \tan A}  \times  \frac{ - 1 +  \tan A }{1 +  \tan A}

= \frac{ - 1 +  \tan A }{1  -   \tan A} \times \frac{  1  -   \tan A }{1 +  \tan A}

\boxed{=  - 1}

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