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

Then prove that:

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

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    What is the source of the promising problem? – 2017-01-12
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    Well i am a Class 10 student and i got it from a pratice book of DR Shimkhada Opt. Maths. – 2017-01-12
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    $$3(\cos A+\cos B)=4(\sin A+\sin B)$$ 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 – 2017-01-12
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    Could u please elaborate the last step. – 2017-01-12
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    $$\dfrac{\sin x}a=\dfrac{\cos x}b=\pm\sqrt{\dfrac{\sin^2x+\cos^2x}{a^2+b^2}}=?$$ – 2017-01-12
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    can u prove that? – 2017-01-12
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    @MyGlasses What's gone? – 2017-01-12
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    as @labbhattacharjee said you must be missing something – 2017-01-12
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    Can u guess what might be missing? – 2017-01-12
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    @Amar30657 The given numbers in your problem are not match with below solution. you say $\cos\frac{A-B}{2}=\frac{5}{24}$ so $A\geq154$. – 2017-01-12
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    Show what you have tried? – 2017-01-12
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    This is not a homework question site. – 2017-01-12
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    @NeWtoN Its not your father's site so dont be oversmart i didnt got the solution so i asked. – 2017-01-13
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    @Amar, Change your tone. Here are not your servants to post you a solution. – 2017-01-13

1 Answers 1

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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 $$\tan{\dfrac{A+B}{2}}=\dfrac{3}{4}$$

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    Well where is the proof? – 2017-01-12
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    so your book is wrong – 2017-01-12
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    I dont think so. But Could u verify by substituting A and B with constants. – 2017-01-12