Question

If X=(1+tan1). (1+tan2).(1+tan3).....(1+tan25)
And Y = (1-tan136).(1+tan137).......(1-tan160)
then XY is
Please answer me quickly ​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

=(1+tan 0o)(1 + tan1°) (1 + tan2°) (1 + tan3°) ………. (1 + tan 45°) {Since , tan 0o = 0}

Now , we know that

tan(A + B) = [ tan(A) + tan(B) ] / [ 1 - tan(A) tan(B) ]

And tan(45) = 1

So

tan(45) = [ tan1 + tan44 ] / [ 1 - tan1 tan44 ]

tan(45) = [ tan2 + tan43 ] / [ 1 - tan2 tan43 ]

.

.

.

tan(45) = [ tan22 + tan23 ] / [ 1 - tan22 tan23 ]

So we have ,

tan1 + tan44 = tan(45) [ 1 - tan1 tan44 ]

tan2 + tan43 = tan(45) [ 1 - tan2 tan43 ]

.

.

.

tan22 + tan23 = tan(45) [ 1 - tan22 tan23 ]

So finally we have ,

(1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + tan45)

= (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + 1)

= 2 (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44)

= 2 (1 + tan1) (1 + tan44) * (1 + tan2) (1 + tan43) * (1 + tan3) (1 + tan42) * ... * (1 + tan22) (1 + tan23)

= 2 (tan1 + tan44 + tan1 tan44 + 1) (tan2 + tan43 + tan2 tan43 + 1) ... (tan22 + tan23 + tan22 tan23 + 1)

= 2 ((tan45)(1 - tan1 tan44) + tan1 tan44 + 1) ((tan45)(1 - tan2 tan43) + tan2 tan43 + 1) ... ((tan45)(1 - tan22 tan23) + tan22 tan23 + 1)

= 2 (1 - tan1 tan44 + tan1 tan44 + 1) (1 - tan2 tan43 + tan2 tan43 + 1) ... (1 - tan22 tan23 + tan22 tan23 + 1)

= 2 (2) (2) ... (2)

= 2 (2^22)

= 2^23