Prove that sin - - - - 2sin- - sin - - -cos - - - - 2cos- - cos - - - - tan-

L.H.S = sin( θ #160; + Φ) + sin( θ #160; Φ) 2sinθcos( θ #160; + Φ) + cos( θ #160; Φ) 2cosθ #160; #160; #160; #160; #160; ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

Prove that s...

Question

Prove that
$\frac{sin (θ + ϕ) - 2 sin θ + sin (θ - ϕ)}{cos (θ + ϕ) - 2 cos θ + cos (θ - ϕ)} = tan θ$ .

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Solution

$L . H . S = \frac{sin (θ + Φ) + sin (θ - Φ) - 2 sin θ}{cos (θ + Φ) + cos (θ - Φ) - 2 cos θ}$

$= \frac{2 sin θ cos Φ - 2 sin θ}{2 cos θ cos Φ - 2 cos θ}$

$= \frac{2 sin θ (cos Φ - 1)}{2 cos θ (cos Φ - 1)}$

$= tan θ = R . H . S$

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8. Prove that :
(i) $\frac{s i n A + s i n 3 A + s i n 5 A}{c o s A + c o s 3 A + c o s 5 A} = t a n 3 A$
(ii) $(i i) \frac{c o s 3 A + 2 c o s 5 A + c o s 7 A}{c o s A + 2 c o s 3 A + c o s 5 A} = \frac{c o s 5 A}{c o s 3 A}$
(iii) $\frac{c o s 4 A + c o s 3 A + c o s 2 A}{c o s 4 A + s i n 3 A + s i n 2 A} = c o t 3 A$
(iv) $\frac{s i n 3 A + s i n 5 A + s i n 7 A + s i n 9 A}{c o s 3 A + c o s 5 A + c o s 7 A + c o s 9 A} = t a n 6 A$
(v) $\frac{s i n 5 A - s i n 7 A + s i n 8 A - s i n 4 A}{c o s 4 A + c o s 7 A + c o s 7 A - c o s 5 A - c o s 8 A} = c o t 6 A$
(vi) $\frac{s i n 5 A + c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A} = t a n A$
(vii) $\frac{s i n 11 A + s i n A + s i n 7 A + s i n 3 A}{c o s 11 A s i n A + c o s 7 A s i n 3 A} = t a n 8 A$
(viii) $\frac{s i n 3 A c o s 4 A - s i n A c o s 2 A}{s i n 4 A s i n A + c o s 6 A c o s A} = t a n 2 A$
(ix) $\frac{s i n A s i n 2 A + s i n 3 A s i n 6 A}{s i n A c o s 2 A + c o s 3 A c o s 6 A} = t a n 5 A$
(x) $\frac{s i n A + 2 s i n 3 A + s i n 5 A}{s i n 3 A + 2 s i n 5 A + s i n 7 A} = \frac{s i n 3 A}{s i n 5 A}$
(xi) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

If sin $θ = \frac{3}{5}$ and cos $ϕ = \frac{- 12}{13}$ where $θ$ and $ϕ$ both lie in the second quadrant, find the values of

(i) sin $(θ - ϕ)$ , (ii) cos $(θ + ϕ)$ , (iii) tan $(θ - ϕ)$ .

Q. Solve the equations

sin (θ - ϕ) = \frac{1}{2}

, and

cos (θ + ϕ) = \frac{1}{2}

Q. Show that

\frac{cos θ + cos ϕ}{sin θ + sin ϕ} = cot (\frac{θ + ϕ}{2})

Q. If

cos θ - sin θ = \sqrt{2} sin θ

, prove that

cos θ + sin θ = \sqrt{2} cos θ

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Prove thatsin(θ+ϕ)−2sinθ+sin(θ−ϕ)cos(θ+ϕ)−2cosθ+cos(θ−ϕ)=tanθ.

L.H.S=sin(θ+Φ)+sin(θ−Φ)−2sinθcos(θ+Φ)+cos(θ−Φ)−2cosθ =2sinθcosΦ−2sinθ2cosθcosΦ−2cosθ =2sinθ(cosΦ−1)2cosθ(cosΦ−1) =tanθ=R.H.S

Prove that
$\frac{sin (θ + ϕ) - 2 sin θ + sin (θ - ϕ)}{cos (θ + ϕ) - 2 cos θ + cos (θ - ϕ)} = tan θ$ .

$L . H . S = \frac{sin (θ + Φ) + sin (θ - Φ) - 2 sin θ}{cos (θ + Φ) + cos (θ - Φ) - 2 cos θ}$

$= \frac{2 sin θ cos Φ - 2 sin θ}{2 cos θ cos Φ - 2 cos θ}$

$= \frac{2 sin θ (cos Φ - 1)}{2 cos θ (cos Φ - 1)}$

$= tan θ = R . H . S$