Question

if tan^2A=1+2tan^2B prove that cos^2B=2cos^2A
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

★ Given:

• tan²A = 1 + 2 tan²B

★ To prove:

➳ cos²B = 2 cos²A

★ Solution:

$\sf{tan^{2} A = 1+ 2tan^{2} B}\\\\\\\Rightarrow \sf{\dfrac{sin^{2} A}{cos^{2} A} = 1 + \dfrac{2 \: sin^{2} B}{cos^{2} B}}$

$\rm{Tanking \: LCM \: on \: the \: right \: hand \: side,}$

$\sf{\dfrac{sin^{2} A}{cos^{2} A} = \dfrac{cos^{2} B + 2sin^{2} B}{cos^{2} B}}$

$\rm{On \: cross \: multiplication,}$

$\sf{sin^{2} A \: (cos^{2}B) = cos^{2}A \: (cos^{2}B+2sin^{2}B)}\\\\\\\Rightarrow \sf{sin^{2}A \: (1 - sin^{2}B) = (1 - sin^{2}A)(1 - sin^{2}B + 2 sin^{2}B)}\\\\\\\Rightarrow \sf{sin^{2}A - sin^{2}A .sin^{2}B = (1 - sin^{2}A)(1 + sin^{2}B)}$

$\Rightarrow \sf{sin^{2}A - sin^{2}A.sin^{2}B = 1 - sin^{2}A + sin^{2}B - sin^{2}A.sin^{2}B}$

Cancel -sin²A.sin²B on both sides.

$\longrightarrow \: \sf{sin^{2} A = 1-sin^{2} A+sin^{2} B}\\\\\\\Rightarrow \sf{2 sin^{2} A = 1 + sin^{2} B}$

Subtracting 2 from both sides,

2 - 2sin²A = 2 - (1 + sin²B)

⇒ 2 (1 - sin²A) = 2 - 1 - sin²B

⇒ 2 (cos²A) = 1 - sin²B

⇒ 2 cos²A = cos²B

∴ cos²B = 2 cos²A