e करना:(i) tand+2tan20 + 4 cot4d = cotf

1.	e करना:(i) tand+2tan20 + 4 cot4d = cotf
Answer» Using the following T- functions of sum & difference of angles, we can prove it. (1) tan 2A = (2tanA) / (1- tan²A) So, (2) tan 4A = ( 2tan2A) / (1- tan² 2A) LHS = tanA + 2tan2A + 4/(tan4A) = tanA + 2tan2A + {4/ (2tan2A)/(1-tan²2A)}….by using (2)nd function = tanA + 2tan2A + 4(1-tan²2A) / 2tan2A = tanA + 2tan2A + 2(1-tan²2A) / tan2A = tanA + {( 2tan² 2A + 2 - 2tan²2A)} / tan2A = tanA + { ( 2/ tan2A ) } = tanA + [2 / {2tanA/(1-tan² A)}]…. by using 1st function = tanA + [ {2( 1-tan²A)} /2tanA ] = {(2tan² A + 2 - 2tan² A )} / 2tanA = 2/ (2 tanA) = 1/tanA = cot A = RHS [ Hence Proved]

Answer»

Using the following T- functions of sum & difference of angles, we can prove it.

(1) tan 2A = (2tanA) / (1- tan²A)

So, (2) tan 4A = ( 2tan2A) / (1- tan² 2A)

LHS = tanA + 2tan2A + 4/(tan4A)

= tanA + 2tan2A + {4/ (2tan2A)/(1-tan²2A)}….by using (2)nd function

= tanA + 2tan2A + 4(1-tan²2A) / 2tan2A

= tanA + 2tan2A + 2(1-tan²2A) / tan2A

= tanA + {( 2tan² 2A + 2 - 2tan²2A)} / tan2A

= tanA + { ( 2/ tan2A ) }

= tanA + [2 / {2tanA/(1-tan² A)}]…. by using 1st function

= tanA + [ {2( 1-tan²A)} /2tanA ]

= {(2tan² A + 2 - 2tan² A )} / 2tanA

= 2/ (2 tanA)

= 1/tanA

= cot A = RHS

[ Hence Proved]

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