Find the value of sin- if a1sin- -b1cos- -0 and a2sin- -b2cos- -c2-0-

Solving the given equation simultaneously, we get sinθ/b_1c_2 b_2c_1=cosθ/c_1a_2 c_2a_1=1/a_1b_2 a_2b_1 sinθ =b_1c_2 b_2c_1/a_1b_2 a_2b_1 ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

Find the valu...

Question

Find the value of $sin θ$ , if $a_{1} sin θ + b_{1} cos θ = 0$ and $a_{2} sin θ + b_{2} cos θ + c_{2} = 0$ .

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a^{2} + b^{2} + c^{2} = 1

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + (b^{2} + c^{2}) c o s θ & b a (1 - c o s θ) & c a (1 - c o s θ) a b (1 - c o s θ) & b^{2} (c^{2} + a^{2}) c o s θ & c b (1 - c o s θ) a c (1 - c o s θ) & b c (1 - c o s θ) & c^{2} + (a^{2} + b^{2}) c o s θ \end{matrix} ∣ ∣ ∣ ∣ ∣

sin θ + 2 cos θ = 1

2 sin θ - cos θ

sin θ \geq 0

a^{2} {sec}^{2} θ - b^{2} {tan}^{2} θ = c^{2} .

sin θ

sec θ + tan θ = 2,

sin θ

0^{\circ} < 0 < 90^{\circ}

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