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I am trying to solve the following trigonometric equation: $$\frac{\cot\theta+\csc\theta}{\tan\theta+\sec\theta}=\cot(\pi/4+\theta/2)\cot \theta/2$$ but unfortunately, after giving a lot of effort I am unable to solve the problem, I think I am missing some obvious steps so the problem is getting messed up but I wasn't able to point out my fault, I also tried to solve it by starting from RHS but I also failed in there.

I will be very much grateful if someone helps me.

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  • $\begingroup$ Have you tried converting everything into $ \sin \theta, \cos \theta$ and simplified the expression? $\endgroup$
    – Calvin Lin
    Nov 20 '20 at 17:49
  • $\begingroup$ yes sir I do, primarily tried to use submultiple angle formula but I wasn't able to prove the result. $\endgroup$ Nov 20 '20 at 17:51
  • $\begingroup$ Can you show what you get when converting into sin, cos, expanding the multiple angle, and then simplifying? $\endgroup$
    – Calvin Lin
    Nov 20 '20 at 17:53
  • $\begingroup$ Actually I wished to write the whole thing I found after reading the guideline but actually as a high school student I am not quite good at latex. ok, after computing a bit I easily found $cot(\theta/2)$ but the remaining expression I get was $\frac{cos\theta}{sin\theta+1}$ from there I didn't able to make any progress. $\endgroup$ Nov 20 '20 at 18:00
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$$\frac{\cot \theta+\csc \theta}{\tan \theta+ \sec \theta}= \frac{1+\cos \theta}{1+\sin \theta}~\frac{\cos \theta}{\sin \theta}$$ $$=\frac{2\cos^2(\theta/2)}{[\sin(\theta/2)+\cos(\theta/2)]^2}\frac{\cos^2(\theta/2)-\sin^2(\theta/2)}{2\sin(\theta/2) \cos(\theta/2)}=\cot(\theta/2) \frac{\cos (\theta/2)-\sin(\theta/2)}{\cos (\theta/2)+\sin(\theta/2)}$$ $$=\cot(\theta/2) \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}=\cot(\theta/2) \tan(\pi/4-\theta/2)=\cot(\theta/2) \cot(\pi/4+\theta/2).$$ Use $\tan(\pi/2-z)=\cot z$.

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Let $c=\cot\dfrac\theta2$

Use Weierstrass substitution

$$\tan\theta=\cdots=\dfrac{2c}{c^2-1}$$

$$\sin\theta=\cdots=\dfrac{2c}{c^2+1}$$

$$\cos\theta=\cdots=\dfrac{c^2-1}{c^2+1}$$

If $c\ne0,$

LHS $=\dfrac{\dfrac{c^2-1}{2c}+\dfrac{c^2+1}{2c}}{\dfrac{2c}{c^2-1}+\dfrac{c^2+1}{c^2-1}}=\dfrac{c(c^2-1)}{(c+1)^2}=c\cdot\dfrac{c-1}{c+1}$ if $c+1\ne0$

Finally use Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$

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