If 2 tan-2-tan-2- prove that cos-3-5 cos-5-3 cos-

We have, 2 tan =α/2=tan β/2 ⇒ tanα/2/tan β/2=1/2 Let tan α/2=H and tan β/2=2K, Then, cos α =1 tan^2 α/2/1+tan^2 α/2=1 K^2/1+K^2 . . . (A0 Also, 3+5 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sin2A and Cos2A in Terms of tanA

If 2 tanα/2=t...

Question

If $2 t a n \frac{α}{2} = t a n \frac{β}{2}$ , prove that $c o s α = \frac{3 + 5 c o s β}{5 + 3 c o s β}$ .

Open in App

Solution

We have,

$2 t a n = \frac{α}{2} = t a n \frac{β}{2} \Rightarrow \frac{t a n \frac{α}{2}}{t a n \frac{β}{2}} = \frac{1}{2}$

Let $t a n \frac{α}{2} = H a n d t a n \frac{β}{2} = 2 K$ ,

Then,

$c o s α = \frac{1 - t a n^{2} \frac{α}{2}}{1 + t a n^{2} \frac{α}{2}} = \frac{1 - K^{2}}{1 + K^{2}}$ . . . (A0

Also,

$\frac{3 + 5 c o s β}{5 + 3 c o s β} = \frac{3 + 5 ⎛ ⎜ ⎝ \frac{1 - t a n^{2} \frac{β}{2}}{1 + t a n^{2} \frac{β}{2}} ⎞ ⎟ ⎠}{5 + 3 ⎛ ⎜ ⎝ \frac{1 - t a n^{2} \frac{β}{2}}{1 + t a n^{2} \frac{β}{2}} ⎞ ⎟ ⎠}$

$= \frac{3 + 5 (\frac{1 - 4 K^{2}}{1 + 4 K^{2}})}{5 + 3 (\frac{1 - 4 K^{2}}{1 + 4 K^{2}})}$
$= \frac{8 - 8 K^{2}}{8 + 8 K^{2}} = \frac{1 - K^{2}}{1 + K^{2}}$ . . . (B)

form (A) and (B)

$c o s α = \frac{3 + 5 c o s β}{5 + 3 c o s β}$

Suggest Corrections

Similar questions

If $2 t a n \frac{α}{2} = t a n \frac{β}{2},$ prove that $c o s α = \frac{3 + 5 c o s β}{5 + 3 c o s β} .$ A.so, determine the value of $c o s α$ when $β = 60^{\circ} .$