Question

prove that
$tan \alpha \times \sin \alpha = \sqrt{ \tan^{2} \alpha - \sin {}^{2} \alpha}$
​

Answer 1

||✪✪ QUESTION ✪✪||

Prove that :- tanA * sinA = √(tan²A - sin²A) .

|| ★★ FORMULA USED ★★ || :-

• TanA = SinA/cosA
• (1 - cos²A) = sin²A .

|| ✰✰ ANSWER ✰✰ ||

Solving LHS First, by putting TanA = SinA/cosA , we get,

→ (SinA/cosA) * sinA

→ (sin²A/cosA)

____________________________

Now, putting This value in Question we get,

→ (sin²A/cosA) = √(tan²A - sin²A)

Squaring both sides now, we get,

→ [ (sin²A/cosA) ]² = (tan²A - sin²A)

→ (sin⁴A/cos²A) = [ (sin²A/cos²A) - sin²A ]

Taking LCM in RHS now,

→ (sin⁴A/cos²A) = [ (sin²A - cos²A * sin²A) /cos²A ]

Taking sin²A common From RHS Numerator now,

→ (sin⁴A/cos²A) = [ sin²A(1-cos²A) /cos²A ]

Putting (1-cos²A = sin²A in RHS Numerator now ,

→ (sin⁴A/cos²A) = [ sin²A * sin²A / cos²A ]

→ (sin⁴A/cos²A) = (sin⁴A/cos²A)

##✪✪ Hence Proved ✪✪##

Answer 2

We Have :-

$tan\alpha * sin\alpha = \sqrt{tan^{2}\alpha-sin^{2}\alpha }$

To Prove :-

$tan\alpha * sin\alpha = \sqrt{tan^{2}\alpha-sin^{2}\alpha }$

Formula Used :-

$sin^{2}x = 1- cos^{2}x\\\\tanx = \frac{sinx}{cosx}$

Solution :-

There are 2 methods to do this :-

1 method :-

$tan\alpha * sin\alpha = \sqrt{tan^{2}\alpha-sin^{2}\alpha }\\\\\frac{sin\alpha }{cos\alpha } * sin\alpha = \sqrt{tan^{2}\alpha-sin^{2}\alpha } \\\\\frac{sin^{2}\alpha }{cos\alpha } = \sqrt{tan^{2}\alpha-sin^{2}\alpha }\\\\Squaring\ both\ sides\\\\\frac{sin^{4}\alpha }{cos^{2}\alpha } = [ tan^{2}\alpha - sin^{2}\alpha ]\\\\\frac{sin^{4}\alpha }{cos^{2}\alpha } = [ \frac{sin^{2}\alpha }{cos^{2}\alpha} - sin^{2}\alpha ]$

$\frac{sin^{4}\alpha}{cos^{2}\alpha}= [ \frac{sin^{2}\alpha-cos^{2}\alpha*sin^{2}\alpha }{cos^{2}\alpha} ]\\\\\frac{sin^{4}\alpha}{cos^{2}\alpha}= [ \frac{sin^{2}\alpha ( 1 - cos^{2}\alpha)}{cos^{2}\alpha} ]\\\\Putting\ 1 -cos^{2}\alpha = sin^{2}\alpha\\\\\frac{sin^{4}\alpha}{cos^{2}\alpha} = [ \frac{sin^{4}\alpha}{cos^{2}\alpha}]\\\\LHS = RHS\\\\Hence\ Proved$

2 Method :-

$Taking\ RHS\\\\\sqrt{tan^{2}\alpha-sin^{2}\alpha} \\\\\sqrt{\frac{sin^{2}\alpha}{cos^{2}\alpha}-sin^{2}\alpha }\\ \\\sqrt{\frac{sin^{2}\alpha-cos^{2}\alpha*sin^{2}\alpha}{cos^{2}\alpha} }\\ \\\sqrt{\frac{sin^{2}\alpha(1-cos^{2}\alpha)}{cos^{2}\alpha} } \\\\\frac{sin\alpha\sqrt{1-cos^{2}\alpha} }{cos\alpha} \\\\Putting\ 1-cos^{2}\alpha = sin^{2}\alpha\\\\\frac{sin^{2}\alpha}{cos\alpha}$

$\frac{sin\alpha}{cos\alpha} *sin\alpha\\\\tan\alpha*sin\alpha\\\\LHS = RHS\\\\Hence\ Proved$