Question

12. Prove that: sin3 θ−cos3θ
sinθ−cosθ
+
sin3 θ+cos3θ
sinθ+cosθ
= 2​

Answer 1

Answer:

Simplify left hand side of sinθ- 2sin3θ2cos3θ-cosθ = tanθ = sinθ (1 - 2sin2θ )cosθ (2cos2θ- 1 ) = sinθ (1 - 2sin2θ )cosθ (2 (1 - sin2θ ) ...

Answer 2

Answer:

$e\pi \sin( \sin( \beta \pi\%e \alpha \csc(\%e\p\pi \sec( \sec( \csc( \csc(\pi\pi \sin(?) ) ) ) ) ) ) )$

$e \sin( \cos( \cos(\pi\pi\pi \sin( \cos(\pi\pi\pi\pe \sin( \sec( \sec(?) ) ) ) ) ) ) )$