Question

if,
tan²A= 1+ 2 tan²B.
Then prove, 2sin²A = 1 +sin^2 B
​

Answer 1

Step-by-step explanation :

  Given :

tan²A = 1 + 2tan²B

  To prove :

2sin²A = 1 + sin²B

  Answer :

     $tan^2A=1+2tan^2B \\\\\\ \frac{sin^2A}{cos^2A} =1+2\frac{sin^2B}{cos^2B} \\\\\\ \frac{sin^2A}{cos^2A} =\frac{cos^2B+2sin^2B}{cos^2B} \\\\\\ sin^2A.cos^2B =cos^2A(cos^2B+2sin^2B) \\\\\\ sin^2A(1-sin^2B)=(1-sin^2A)(1-sin^2B+2sin^2B) \\\\\\ sin^2A-sin^2Asin^2B = (1-sin^2A)(1+sin^2B) \\\\\\ sin^2A-sin^2A.sin^2B=1(1+sin^2B)-sin^2A(1+sin^2B) \\\\\\ sin^2A-sin^2A.sin^2B =1+sin^2B-sin^2A-sin^2A.sin^2B \\\\\\ sin^2A=1+sin^2B-sin^2A \\\\\\ sin^2A+sin^2A=1+sin^2B \\\\\\ \boxed{2sin^2A=1+sin^2B}$

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$\bf \longrightarrow tan \theta =\frac{sin\theta}{cos\theta} \\\\\\ \longrightarrow sin^2\theta+cos^2\theta=1 \\ => cos^2\theta=1-sin^2\theta$

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Some identities :

• sin² A + cos² A = 1

• 1+tan² A = sec² A

• 1+cot² A = cosec² A

• sin (x+y) = sin(x)cos(y) + cos(x)sin(y)

• cos(x+y) = cos(x)cos(y) – sin(x)sin(y)  

• tan(x+y) = (tan x + tan y) / [ (1−tan x)(tan y)  ]
• sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y) / [ (1+tan x)( tan y) ]