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**wintersoltice** 1.

Given that $\displaystyle \frac{cos(A-B)}{cos(A+B)}$ = $\displaystyle \frac{7}{5}$, prove that cos A cos B = 6 sin A sin B

Mr F says: Substitute from the appropriate double angle formula in the numerator and denominator, cross multiply to get rid of fractions, and simplify.

and deduce a relationship between tan A and tan B.

Mr F says: Divide both sides of cos A cos B = 6 sin A sin B by cos A cos B .....

Given further that A+B=$\displaystyle 45^\circ$, calculate the value of tan A + tan B.

Mr F says: $\displaystyle {\color{red}\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}}$. And from the previous result you know what tan A tan B is equal to.

2.

If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.

Mr F says: Here's a start: $\displaystyle {\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C }$ $\displaystyle {\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }$ ......

Hence, prove that sin A - sin(B-C)= 2 cos B sin C .

Mr F says: It's not difficult to use the compound angle formulae to show that sin (B + C) = sin (B - C) + 2 sin C cos B ......