# Trigonometry Prove that :

If $\frac{3}{\sec A}+ \frac{3}{\sec B} = \frac{4}{\csc A} + \frac{4}{\csc B}$

Then prove that:

$24\cos{\frac{(A-B)}{2}} = \pm 5$

• What is the source of the promising problem? – lab bhattacharjee Jan 12 '17 at 11:34
• Well i am a Class 10 student and i got it from a pratice book of DR Shimkhada Opt. Maths. – Amar30657 Jan 12 '17 at 11:35
• $$3(\cos A+\cos B)=4(\sin A+\sin B)$$ Using Prosthaphaeresis Formulas, $$6\cos\dfrac{A+B}2\cos\dfrac{A-B}2=8\sin\dfrac{A+B}2\cos\dfrac{A-B}2$$ If $\cos\dfrac{A-B}2\ne0,$ $$\dfrac{\sin\dfrac{A+B}2}3=\dfrac{\cos\dfrac{A+B}2}4=\pm\dfrac1{\sqrt{3^2+4^2}}$$ So, we must be missing something – lab bhattacharjee Jan 12 '17 at 11:37
• Could u please elaborate the last step. – Amar30657 Jan 12 '17 at 11:40
• $$\dfrac{\sin x}a=\dfrac{\cos x}b=\pm\sqrt{\dfrac{\sin^2x+\cos^2x}{a^2+b^2}}=?$$ – lab bhattacharjee Jan 12 '17 at 11:43

since $$3(\cos{A}+\cos{B})=4(\sin{A}+\sin{B})$$ so we havce $$6\cos{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}=8\sin{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}$$so we have $$\cos{\dfrac{A-B}{2}}=0$$ or $$\tan{\dfrac{A+B}{2}}=\dfrac{3}{4}$$