Sin(A+B)+sin(A+B)/Cos(A+B)+cos(A-B)=TanA

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Solution

Expand the given formula by Sin(A+B)=sinAcosB+cosAsinB
sin(A-B)= sinAcosB-cosAsinB
cos (A-B)=cosAcosB+sinAsinB
cos (A+B)=cosAcosB-sinAsinB
=[(sinAcosB+cosAsinB) + (sinAcosB-cosAsinB)] /[(cosAcosB-sinAsinB) + (cosAcosB+sinAsinB)]

=2sinAcosB / 2cosAcosB

=sinA/cosA= tanA
hence proved

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Similar questions

Q. Assertion :

\frac{sin (A + B) + sin (A - B)}{cos (A + B) + cos (A - B)} = tan A

Reason:

sin (A + B) + sin (A - B) = sin A

and

cos (A + B) + cos (A - B) = cos A

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

If $\frac{m}{n} = \frac{t a n A}{t a n B}$ , find the value of $\frac{m + n}{m - n}$

If cos(A-B) = 3/5 and tanA*tanB = 2 then

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Q. Prove that

\frac{cos (A + B + C) + cos (- A + B + C) + cos (A - B + C) + cos (A + B - C)}{sin (A + B + C) + sin (- A + B + C) - sin (A - B + C) + sin (A + B - C)} = cot B

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Sin(A+B)+sin(A+B)/Cos(A+B)+cos(A-B)=TanA

Expand the given formula by Sin(A+B)=sinAcosB+cosAsinB sin(A-B)= sinAcosB-cosAsinB cos (A-B)=cosAcosB+sinAsinB cos (A+B)=cosAcosB-sinAsinB =[(sinAcosB+cosAsinB) + (sinAcosB-cosAsinB)] /[(cosAcosB-sinAsinB) + (cosAcosB+sinAsinB)] =2sinAcosB / 2cosAcosB =sinA/cosA= tanA hence proved

Expand the given formula by Sin(A+B)=sinAcosB+cosAsinB
sin(A-B)= sinAcosB-cosAsinB
cos (A-B)=cosAcosB+sinAsinB
cos (A+B)=cosAcosB-sinAsinB
=[(sinAcosB+cosAsinB) + (sinAcosB-cosAsinB)] /[(cosAcosB-sinAsinB) + (cosAcosB+sinAsinB)]

=2sinAcosB / 2cosAcosB

=sinA/cosA= tanA
hence proved