Question

prove that
( sinθ-2sin3θ) /(2cos3θ-cosθ ) =tanθ


(PLEASE GIVE ANSWER IF YOU KNOW FAST)​

Answer 1

GIVEN :-

( taking ∅ = a )

$\implies \rm{ \dfrac{sin \: a - 2 \: {sin}^{3} a}{2 \: cos {}^{3} a - cos \: a} }$

TO FIND :-

$\implies \rm{ \dfrac{sin \: a - 2 \: {sin}^{3} a}{2 \: cos {}^{3} a - cos \: a} = tan \: a }$

SOLUTION :-

taking LHS :-

$\implies \rm{ \dfrac{sin \: a - 2 \: {sin}^{3} a}{2 \: cos {}^{3} a - cos \: a} }$

taking sin a common

$\implies \rm{ \dfrac{sin \: a(1 - 2 \: {sin}^{2} a)}{2 \: cos {}^{3} a - cos \: a} }$

taking cos a common

$\implies \rm{ \dfrac{sin \: a(1 - 2 \: {sin}^{2} a)}{cos \: a( 2 \: {cos}^{2} a -1 )} }$

now we know that according to trignomatric ratios :-

$\implies \boxed{ \rm{tan \: a = \dfrac{sin \: a}{cos \: a} }}$

hence ,

$\implies \rm{ \dfrac{sin \: a(1 - 2 \: {sin}^{2} a)}{cos \: a( 2 \: {cos}^{2} a -1 )} } = \rm{ (tan \: a)\dfrac{(1 - 2 \: {sin}^{2} a)}{ ( 2 \: {cos}^{2} a -1 )} }$

now according to trignomatric identities :-

$\implies \boxed{ \rm{ {cos}^{2} a = 1 - {sin}^{2} a}}$

so ,

$\implies\rm{ (tan \: a)\dfrac{(1 - 2 \: {sin}^{2} a)}{ ( 2 (1 - {sin}^{2} a )-1 )} }$

now solving further ( denominator )

$\implies\rm{ (tan \: a)\dfrac{(1 - 2 \: {sin}^{2} a)}{ ( 2 - 2 \: {sin}^{2} a - 1)} }$

$\implies\rm{ (tan \: a)\dfrac{(1 - 2 \: {sin}^{2} a)}{ ( 1 - 2 \: {sin}^{2} a )} }$

now cancel outting (1 - 2 sin² a)

$\implies\rm{ \bf{tan \: a} }$

so , LHS = RHS

and

$\implies \boxed{ \boxed {\rm{ \dfrac{sin \: a - 2 \: {sin}^{3} a}{2 \: cos {}^{3} a - cos \: a} = tan \: a }}}$

OTHER INFORMATION :-

TRIGNOMETRIC IDENTITIES

• sin²∅ + cos²∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

TRIGNOMETRIC RATIOS

• sin ∅ = 1 / cosec ∅

• cos ∅ = 1 / sec ∅

• tan ∅ = 1 / cot ∅

TRIGNOMETRIC COMPLEMENTRY ANGLES

• sin ∅ = cos ( 90 - ∅ )

• cos ∅ = sin ( 90 - ∅ )

• sec ∅ = cosec ( 90 - ∅ )

• cosec ∅ = sec ( 90 - ∅ )

• tan ∅ = cot ( 90 - ∅ )

• cot ∅ = tan ( 90 - ∅ )