Question

show that cos2A/secA-sin2A/cosecA=cos3A

Answer 1

• cos2A/1/cosA-sin2A/1/sinA=cos3A
• cos2A*cosA-sin2A*sinA
• cos(2A+A)
• cos3A=cos3A
• L.H.S=R.H.S
• hence it is proved

Answer 2

Question :-

Show that

$\rm \: \dfrac{cos2A}{secA} - \dfrac{sin2A}{cosecA} \: = \: cos3A$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \dfrac{cos2A}{secA} - \dfrac{sin2A}{cosecA}$

can be rewritten as

$\rm \: = \:cos2A \: cosA \: - \: sin2A \: sinA$

We know,

$\boxed{\sf{  \:\rm \: cosx \: cosy \: - \: sinx \: siny \: = \: cos(x + y) \: \: }}$

So, using this identity, we get

$\rm \: = \:cos(2A + A)$

$\rm \: = \:cos 3A$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \dfrac{cos2A}{secA} - \dfrac{sin2A}{cosecA} \: = \: cos3A \: \: }}$

$\rule{190pt}{2pt}$

Alternative Method :-

Consider LHS

$\rm \: \dfrac{cos2A}{secA} - \dfrac{sin2A}{cosecA}$

can be further rewritten as

$\rm \:= \: \dfrac{2cos^{2} A - 1}{secA} - \dfrac{2sinA \: cosA}{cosecA}$

$\rm \: = \:( {2cos}^{2}A - 1)cosA - 2 {sin}^{2}A \: cosA$

$\rm \: = \: {2cos}^{3}A - cosA - 2 (1 - {cos}^{2}A) \: cosA$

$\rm \: = \: {2cos}^{3}A - cosA - 2cosA + {2cos}^{3}A$

$\rm \: = \: {4cos}^{3}A - 3cosA$

$\rm \: = \:cos3A$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \dfrac{cos2A}{secA} - \dfrac{sin2A}{cosecA} \: = \: cos3A \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{\sf{  \:\rm \: sin2x = 2sinx \: cosx \: \: }}$

$\boxed{\sf{  \:\rm \: cos2x = {2cos}^{2} x - 1 \: \: }}$

$\boxed{\sf{  \:\rm \: cos2x = 1 - {2sin}^{2} x \: \: }}$

$\boxed{\sf{  \:\rm \: cos3x = {4cos}^{3}x - 3cosx \: \: }}$

$\boxed{\sf{  \:\rm \: sin3x =3sinx - {4sin}^{3}x \: \: }}$