Prove the following- sin25O-cos 25o-cos25o-sin25o-sin -

sin(25^∘+θ)cos(25^∘ ϕ) cos(25^∘+θ)sin(25^∘ ϕ) =sin((25^∘+θ (25^∘ ϕ))) =sin(θ+ϕ) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove the fol...

Question

Prove the following:
$sin (25^{O} + θ) \cdot cos (25^{o} - ϕ) - cos (25^{o} + θ) \cdot sin (25^{o} - ϕ) = sin (θ + ϕ)$

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Solution

$s i n (25^{\circ} + θ) c o s (25^{\circ} - ϕ) - c o s (25^{\circ} + θ) s i n (25^{\circ} - ϕ) = s i n ((25^{\circ} + θ - (25^{\circ} - ϕ))) = s i n (θ + ϕ)$

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Similar questions

Q. Prove that

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

sin (25^{0} + θ) . cos (25^{0} - ϕ) - cos (25^{0} + θ) . sin (25^{0} - ϕ) = sin (θ + ϕ)

Q. If

θ + ϕ = α

and

tan θ = k tan ϕ

, then prove that

sin (θ - ϕ) = \frac{k - 1}{k + 1} sin α

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) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

Q. Prove that

{tan}^{4} ϕ + {tan}^{2} ϕ = {sec}^{4} ϕ - {sec}^{2} ϕ

, for

0^{\circ} \leq ϕ \leq 90^{\circ}

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Prove the following:sin(25O+θ)⋅cos(25o−ϕ)−cos(25o+θ)⋅sin(25o−ϕ)=sin(θ+ϕ)

sin(25∘+θ)cos(25∘−ϕ)−cos(25∘+θ)sin(25∘−ϕ)=sin((25∘+θ−(25∘−ϕ)))=sin(θ+ϕ)

Prove the following:
$sin (25^{O} + θ) \cdot cos (25^{o} - ϕ) - cos (25^{o} + θ) \cdot sin (25^{o} - ϕ) = sin (θ + ϕ)$

$s i n (25^{\circ} + θ) c o s (25^{\circ} - ϕ) - c o s (25^{\circ} + θ) s i n (25^{\circ} - ϕ) = s i n ((25^{\circ} + θ - (25^{\circ} - ϕ))) = s i n (θ + ϕ)$