If cos - - -sin - - -0 and tan- 1- then find the value of -

Given, cos( α +β #160;) #160; +sin( α β #160;) #160; =0 ⇒cosα #160;cosβ #160;sinα #160;sinβ #160; +sinα #160;cosβ #160; cosα ...

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Determinant

If cos α +β...

Question

If $cos (α + β) + sin (α - β) = 0$ and $tan β \neq 1,$ then find the value of $α$

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Solution

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cos (α + β) = \frac{4}{5}; sin (α - β) = \frac{5}{13}

α, β

0

\frac{π}{4}

tan 2 α

c o s α + c o s β + c o s γ = s i n α + s i n β + s i n γ = 0

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

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

0, α, β < \frac{π}{4}

cos (α + β) = \frac{4}{5}

sin (α - β) = \frac{5}{13}

tan 2 α =

cos (α + β) = \frac{4}{5}

sin (α - β) = \frac{5}{13}

α, β

0^{\circ}

45^{\circ}

tan 2 α

sin (α + β) = 1

sin (α - β) = \frac{1}{2}

0 \leq α, β \leq \frac{π}{2}

tan (α + 2 β)

tan (2 α + β)

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