-2
$\begingroup$

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

Then prove that:

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

$\endgroup$
  • 1
    $\begingroup$ What is the source of the promising problem? $\endgroup$ – lab bhattacharjee Jan 12 '17 at 11:34
  • $\begingroup$ Well i am a Class 10 student and i got it from a pratice book of DR Shimkhada Opt. Maths. $\endgroup$ – Amar30657 Jan 12 '17 at 11:35
  • $\begingroup$ $$3(\cos A+\cos B)=4(\sin A+\sin B)$$ 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 $\endgroup$ – lab bhattacharjee Jan 12 '17 at 11:37
  • $\begingroup$ Could u please elaborate the last step. $\endgroup$ – Amar30657 Jan 12 '17 at 11:40
  • $\begingroup$ $$\dfrac{\sin x}a=\dfrac{\cos x}b=\pm\sqrt{\dfrac{\sin^2x+\cos^2x}{a^2+b^2}}=?$$ $\endgroup$ – lab bhattacharjee Jan 12 '17 at 11:43
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Well where is the proof? $\endgroup$ – Amar30657 Jan 12 '17 at 11:38
  • $\begingroup$ so your book is wrong $\endgroup$ – math110 Jan 12 '17 at 11:39
  • $\begingroup$ I dont think so. But Could u verify by substituting A and B with constants. $\endgroup$ – Amar30657 Jan 12 '17 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.