Sin 3 A-cos 3 A-sin A-cos A-sin 3 A-cos 3 A-sin A-cos A-2

LHS=sin^3 A +cos^3 A/sin A +cos A+sin^3 A cos^3 A/sin A cos A =(sin A +cos A)(sin^2 A+cos^2 sin A cos A)/(sin A +cos)+(sin A cos A)(sin^2 A+cos^2+sin A cos A)/ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

sin 3 A+cos 3...

Question

$\frac{s i n^{3} A + c o s^{3} A}{s i n A + c o s A} + \frac{s i n^{3} A - c o s^{3} A}{s i n A - c o s A} = 2$

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Solution

$L H S = \frac{s i n^{3} A + c o s^{3} A}{s i n A + c o s A} + \frac{s i n^{3} A - c o s^{3} A}{s i n A - c o s A}$

$= \frac{(s i n A + c o s A) (s i n^{2} A + c o s^{2} - s i n A c o s A)}{(s i n A + c o s)} + \frac{(s i n A - c o s A) (s i n^{2} A + c o s^{2} + s i n A c o s A)}{s i n A c o s}$

(Using $a^{3} + b^{3} = (a + b) (a^{2} + b - 2 a b) a n d a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)]$

$= (1 - s i n A c o s A) + (1 + s i n A c o s A) (∵ s i n^{2} A + c o s^{2} A = 1)$

=2=RHS.

Hence proved.

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Similar questions

Q. Prove that

\frac{S i n^{3} A + c o s^{3} A}{s i n A + c o s A} + \frac{S i n^{3} A - c o s^{3} A}{s i n A - c o s A} = 2

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Prove:

\frac{{c o s}^{3} A + {s i n}^{3} A}{c o s A + s i n A} + \frac{{c o s}^{3} A - {s i n}^{3} A}{c o s A - s i n A} = 2

Find the value of $\frac{c o s^{3} A - c o s 3 A}{c o s A}$ + $\frac{s i n^{3} A - s i n 3 A}{s i n A}$

Q. On simplifying

\frac{{sin}^{3} A + sin 3 A}{sin A} + \frac{{cos}^{3} A - cos 3 A}{cos A}

we get

$\frac{{cos}^{3} A - cos 3 A}{cos A}$ + $\frac{{sin}^{3} A - sin 3 A}{sin A} =$

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sin3A+cos3AsinA+cosA+sin3A−cos3AsinA−cosA=2

LHS=sin3A+cos3AsinA+cosA+sin3A−cos3AsinA−cosA =(sinA+cosA)(sin2A+cos2−sinAcosA)(sinA+cos)+(sinA−cosA)(sin2A+cos2+sinAcosA)sinAcos (Using a3+b3=(a+b)(a2+b−2ab)anda3−b3=(a−b)(a2+b2+ab)] =(1−sinAcosA)+(1+sinAcosA)(∵sin2A+cos2A=1) =2=RHS. Hence proved.

$\frac{s i n^{3} A + c o s^{3} A}{s i n A + c o s A} + \frac{s i n^{3} A - c o s^{3} A}{s i n A - c o s A} = 2$