If tan- - xa - tan- - yb - tan- - xz then prove that a - ba - bsin2 x - y - b - cb - csin2y - z - c - ac - asin2z - x - 0

ab = #10;tan (θ + x)tan (θ + y) Apply compo and #10;divi. #10; ∴ a + ba b = #10;sin (2θ + x + y)sin (x y) ...

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Trigonometric Ratios of Common Angles

If tanθ + ...

Question

If $\frac{t a n (θ + x)}{a} = \frac{t a n (θ + y)}{b} = \frac{t a n (θ + x)}{z}$ then prove that $\frac{a + b}{a - b} s i n^{2} (x - y) + \frac{b + c}{b - c} s i n^{2} (y - z) + \frac{c + a}{c - a} s i n^{2} (z - x) = 0$

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\frac{x}{tan (θ + α)} = \frac{y}{tan (θ + β)} = \frac{z}{tan (θ + y)}

\sum \frac{x + y}{x - y} {sin}^{2} (α - β) = 0

\frac{x}{tan (θ + α)} = \frac{y}{tan (θ + β)} = \frac{z}{tan (θ + γ)}

\frac{x + y}{x - y} {sin}^{2} (α - β) + \frac{y + z}{y - z} {sin}^{2} (β - γ) + \frac{z + x}{z - x} {sin}^{2} (γ - α) = 0

θ = {tan}^{- 1} \sqrt{x} + {tan}^{- 1} \sqrt{y} + {tan}^{- 1} \sqrt{z}

x = \frac{a^{2} (a + b + c)}{a b c}, y = \frac{b^{2} (a + b + c)}{a b c}, z = \frac{c^{2} (a + b + c)}{a b c}

tan θ = 0.

\frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b}

(b - c) x + (c - a) y + (a - b) z = 0

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

\frac{x}{y}

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