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

Have, a tan α = 3 tan β ⇒ tan α/tan β=3/2 Let tan α = 3K and tan β = 2K Now, tan (α β)=tan α tan β/1 + tan α. tan β =3K 2K/1+3K.2K=K/1+6K2 . . .(A) Al ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If 2 tanα=3 t...

Question

If 2 $t a n α = 3 t a n β$ , prove that $t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$

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Solution

Have, $a t a n α = 3 t a n β$

$\Rightarrow \frac{t a n α}{t a n β} = \frac{3}{2}$

Let $t a n α = 3 K a n d t a n β = 2 K$

Now, $t a n (α - β) = \frac{t a n α - t a n β}{1 + t a n α . t a n β}$

$= \frac{3 K - 2 K}{1 + 3 K .2 K} = \frac{K}{1 + 6 K 2}$ . . .(A)

Also,

$\frac{s i n 2 β}{5 - c o s 2 β} = \frac{\frac{2 t a n β}{1 + t a n^{2} β}}{5 - (\frac{1 - t a n^{2} β}{1 + t a n^{2} β})} = \frac{\frac{2.2 K}{1 + 4 K^{2}}}{5 - (\frac{1 - 4 K^{2}}{1 + 4 K^{2}})} = \frac{4 K}{5 + 20 K^{2} - 1 + 4 K^{2}}$
$= \frac{4 K}{4 + 24 K^{2}} = \frac{K}{1 + 6 K^{2}}$ . . . (B)

from (A) and (B)

$t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$

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If 2 tan α=3 tan β, prove that tan(α−β)=sin 2β5−cos2β

If 2 $t a n α = 3 t a n β$ , prove that $t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$