If tan2A=cot2B, Then find the value of sec(A+B).
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Answered by
1
tan2A = cot2B
=tan2A =tan(90-2B) (tan(a) = cot(90-a))--remember this
= 2A = 90-2B
= 2A + 2B = 90
= A+B = 45
therefore, sec(A+B) = sec(45)
sec(45) = √2
=tan2A =tan(90-2B) (tan(a) = cot(90-a))--remember this
= 2A = 90-2B
= 2A + 2B = 90
= A+B = 45
therefore, sec(A+B) = sec(45)
sec(45) = √2
Answered by
0
tan 2a=cot 2b
so tan 45=cot 45
a=b=45
sec a=√²
sec b=√²
so the value is √²+√²
so tan 45=cot 45
a=b=45
sec a=√²
sec b=√²
so the value is √²+√²
anjanik:
i think kanugula gave me the right answer. Ur answer is not right
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