Question

some one help in trigonometry plz
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Answer 1

Answer:

Step-by-step explanation:

(sinθ- 2 sin^3θ)/(2cos^3θ-cosθ)

=sinθ (1-2 sin^2θ) / cosθ (2cos^2θ -1 )

= sin θ [1- 2 (1- cos^2 θ) /cos θ(2cos ^2 θ -1)

= sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1)

= sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1)

= sin θ/ cos θ

= tanθ

see in attachment

plz mark as brainlist

Answer 2

Answer:

divide numerator by sin • (• = theta)

and divide denominator by cos •

now equation will be...

(sin •/ cos •)× (1-2sin^•)/(2cos^2• - 1)

(expand 1 as sin^2•+cos^2•)

or tan • × (cos^2• - sin^2•)/(cos^2• - sin^2•)

=> tan•