If sin -1 and sin -1-2- where 0 - -2- then find the values of tan -2 - and tan 2 -

We have sin(α+ β)=1 ⇒ sin( α β)= sin π/2 ⇒α+ β= π/2 .....(i) and, sin α β= 1/2 ⇒ sin(α β)= sin π/6 ⇒α β= π/6 ....(ii) Adding equations (i) and (ii), we get ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If sin α+β=1 ...

Question

If $s i n (α + β) = 1$ and $s i n (α - β) = \frac{1}{2},$ where $0 \leq α, β \leq \frac{π}{2},$ then find the values of $t a n (α + 2 β)$ and $t a n (2 α + β)$ .

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Solution

We have
$s i n (α + β) = 1$
$\Rightarrow s i n (α - β) = s i n \frac{π}{2}$
$\Rightarrow α + β = \frac{π}{2}$ .....(i)
and, $s i n α - β = \frac{1}{2}$
$\Rightarrow s i n (α - β) = s i n \frac{π}{6}$
$\Rightarrow α - β = \frac{π}{6}$ ....(ii)
Adding equations (i) and (ii), we get
$2 α = \frac{π}{2} + \frac{π}{6} = \frac{4 π}{6} = \frac{2 π}{3}$
$\Rightarrow α = \frac{π}{3}$
Putting $α = \frac{π}{3}$ in equation (i), we get
$\frac{π}{3} + β = \frac{π}{2}$
$\Rightarrow β = \frac{π}{2} - \frac{π}{3}$
$\Rightarrow β = \frac{3 π - 2 π}{6} = \frac{π}{6}$
$\Rightarrow β = \frac{π}{6}$
Now, $t a n (α + 2 β) = t a n (\frac{π}{3} + 2 \times \frac{π}{6})$
$= t a n (\frac{π}{3} + \frac{π}{3})$
$= t a n \frac{2 π}{3}$
$= t a n (\frac{π}{2} + \frac{π}{6})$
$= - c o t \frac{π}{6}$
$= - \sqrt{3}$
$∴ t a n (α + 2 \sqrt{3}) = - \sqrt{3}$
and, $t a n (2 α + β) = t a n (2 \times \frac{π}{3} + \frac{π}{6})$
$= t a n (\frac{2 π}{3} + \frac{π}{6})$
$= t a n (\frac{4 π + π}{6})$
$= t a n (\frac{5 π}{6})$
$= (\frac{π}{2} + \frac{π}{3})$
$= - c o t \frac{π}{3}$
$= \frac{- 1}{\sqrt{3}}$
$∴ t a n (2 α + β) = \frac{- 1}{\sqrt{3}}$

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If sin(α+β)=1 and sin(α−β)=12, where 0≤α,β≤π2, then find the values of tan(α+2β) and tan(2α+β).

If $s i n (α + β) = 1$ and $s i n (α - β) = \frac{1}{2},$ where $0 \leq α, β \leq \frac{π}{2},$ then find the values of $t a n (α + 2 β)$ and $t a n (2 α + β)$ .