If cos - - - - 0- then sin - - - can be reduced to-

The correct option is B cos #160; 2 β cos( α +β #160; ) #160; =0 ⇒α +β =90 #176; ⇒α =90 #176; β ⇒sin( α β #160; ) ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

If cos α +...

Question

If $cos$ $(α + β) = 0$ , then $sin$ $(α - β)$ can be reduced to:

cos

β

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cos

2 β

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sin

α

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sin

2 α

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Solution

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Similar questions

α

β

π < α - β < 2 π

sin α + sin β = - \frac{21}{65}

cos α + cos β = \frac{17}{65}

cos \frac{α - β}{2}

β

α

c o s α + c o s (α + β) + c o s (α + β + γ) = 0

s i n α + s i n (α + β) + s i n (α + β + γ) = 0

β = γ = 2 π / 3

2 s i n^{2} β + 4 c o s (α + β) s i n α s i n β + c o s 2 (α + β) =

If $s i n α, s i n β, c o s α$ are in GP, then roots of equation $x^{2} s i n β + 2 x c o s β + s i n β = 0$ are

\frac{cos α}{cos θ} + \frac{sin α}{sin θ} = \frac{cos β}{cos θ} + \frac{sin β}{sin θ} = 1

α

β

π

\frac{cos α cos β}{{cos}^{2} θ} + \frac{sin α sin β}{{sin}^{2} θ} + 1 = 0

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