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1+tan^2A +(1+1upon tan^2A) = 1upon sin^2A-sin^4A . proof this and how it is sin^4A explain it in detail


jaywadmare 12th Sep, 2020
Answer (1)
anshika.verma2019 13th Nov, 2020

Respected User,

It is given that 1 + tan 2 A + (1 + 1/tan 2 A)

and we have to prove that 1 + tan 2 A + (1 + 1/tan 2 A)  = ______1______

(sin 2 A – sin 4 A)


So, let

1 + tan 2 A + (1 + 1/tan 2 A) _____ eqn. 1

Now, since 1 + tan 2 x = sec 2 x

Therefore, eqn. 1 can be written as

sec 2 A + ((tan 2 A+1)/tan 2 A)

which will further reduce to

sec 2 A + (sec 2 A/tan 2 A)

= sec 2 A + (cos 2 A/cos 2 A*sin 2 A)

= sec 2 A + (1/sin 2 A)

Again, secA = 1/cosA

Thus, rewriting sec 2 A as 1/cos 2 A

= 1/cos 2 A + 1/sin 2 A

Now, take LCM

= sin 2 A + cos 2 A/(sin 2 A*cos 2 A)

Rewrite sin 2 A + cos 2 A = 1 and cos 2 A as 1 – sin 2 A

= 1/ (sin 2 A*(1 – sin 2 A))

Now, multiply the values in denominator.

= 1/ (sin 2 A*1 – (sin 2 A)(sin 2 A))

= 1/(sin 2 A – sin 4 A)

Hope, I was able to solve your query.

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