Question

. Prove that cos2 A-sin2A = 2cos2A -1.​

Answer 1

Property to be used:-

$\underline{\boxed{\sf Sin²Ø = 1-Cos²Ø}}$

Solution

L.H.S :-

$\dashrightarrow$ $\sf Cos²A-Sin²A$

$\dashrightarrow$ $\sf Cos²A-(1-Cos²A)$

$\dashrightarrow$ $\sf Cos²A-1+Cos²A$

$\dashrightarrow$ $\sf 2Cos²A-1$ = R.H.S

hence, proved

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Additional information:-

• $\sf Sec²Ø-Tan²Ø = 1$
• $\sf Cosec²Ø-Cot²Ø = 1$

Answer 2

$\sf{We \: have \: to \: prove,}$

$• \: { \cos }^{2}A - { \sin }^{2} A = 2 { \cos }^{2} A - 1$

$\sf{We \: know \: that, }$

${ \sin}^{2} \theta + { \cos}^{2} \theta = 1$

$\rightarrow { \sin}^{2} \theta =1 - { \cos}^{2} \theta$

$\bf{Left \: Side ={ \cos }^{2}A - { \sin }^{2} A }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = { \cos }^{2}A - (1 - { \cos }^{2}A)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = { \cos }^{2}A - 1 + { \cos }^{2}A$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = 2{ \cos }^{2}A - 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = Right \: Side$

$\red{ \boxed { \therefore \text{Left Side = Right Side}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\text{Hence Proved}}$