Question

please solve it sin (A+B) + cos (A-B)/ sin (A-B) + cos (A+B) = sec 2B + tan 2B​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

Sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B) [By identity] and

Cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)

Adding these two, sin(A+B) + cos(A-B) =

=sin(A)*cos(B) + cos(A)*sin(B) + cos(A)*cos(B) + sin(A)*sin(B)

dividing throughout by cos(A)*cos(B),

sin(A+B) + cos(A-B) = cos(A)*cos(B)[tan(A) + tan(B) + 1 + tan(A)*tan(B)]

Grouping and factorizing,

sin(A+B) + cos(A-B) = cos(A)*cos(B)*{1+tan(A)}*{1 + tan(B)} ------(1)

Similarly proceeding and simplifying,

sin(A-B) + cos(A+B) = cos(A)*cos(B)*{1+tan(A)}*{1 - tan(B)} -------(2)

Dividing (1) by (2),

[sin(A+B)+cos(A-B)/sin(A-B)+cos(A+B)] = {1 + tan(B)}/{1 - tan(B)}

Multiplying numerator and denominator by {1 + tan(B)},

[sin(A+B)+cos(A-B)/sin(A-B)+cos(A+B)] = [1 + tan^2(B) + 2tan(B)]/{1 - tan^2(B)}

Multiplying both numerator and denominator by cos^2(B),

[sin(A+B)+cos(A-B)/sin(A-B)+cos(A+B)] =

=[cos^2(B) + sin^2(B) + 2sin(B)cos(B)]/{cos^2(B) - sin^2(B)}

==> [sin(A+B)+cos(A-B)/sin(A-B)+cos(A+B)] =

= [1 + sin(2B)]/cos(2B) = 1/cos(2B) + sin(2B)/cos(2B) = sec(2B) + tan(2B)

Thus it is proved that

[sin(A+B)+cos(A-B)/sin(A-B)+cos(A+B)] = sec(2B) + tan(2B)