Question

LONG ANSWER QUESTIONS (4 MARKS):
11.If tan A + sin A = m and tanA - sin A = n ,show that ma- n2 = 4 mn​

Answer 1

Correct Question

If tan A + sin A = m and tanA - sin A = n ,show that m²- n² = 4√mn

Answer :

Given :

• m = tan(A) + sin(A)

• n = tan(A) - sin(A)

To Prove

m² - n² = 4√mn

LHS

m² - n²

➠ (m + n)(m - n)

➠ [(tan A + sin A) + (tan A - sin A)][(tanA + sin A) - (tan A - sin A)]

➠ (2tan A)(2 sin A)

➠ 4tan A.sin A______________(1)

RHS

4√mn

➠ 4√(tan A + sin A)(tan A - sin A)

➠ 4√(tan²A - sin²A)

➠ 4√(sin²A/cos²A - sin²A)

➠ 4 × {√(sin²A - sin²Acos²A)/cos²A }

➠ 4 × {√[sin²A(1 - cos²A)]/cos²A }

➠ 4 × √tan²A.sin²A

➠ 4tan A.sin A__________(2)

From equations (1) and (2),

m² - n² = 4√mn

Answer 2

$\sf Correct\: Question :$

If tan A + sin A = m and tan A - sin A = n , then show that m² - n² = 4√mn

$\rule{120}{0.8}$

$\sf Answer :$

⠀

$\longrightarrow\:\:\rm\tan(A)+ \sin(A)=m \\\\\\\longrightarrow\:\:\rm(\tan(A)+ \sin(A))^2=(m)^2\\\\\\\longrightarrow\:\:\rm\tan^2(A)+\sin^2(A)+2\tan(A)\sin(A)=m^2\qquad\sf...eq(l)$

⠀⠀⠀⠀$\rule{100}{1}$

$\longrightarrow\:\:\rm\tan(A) - \sin(A)=n \\\\\\\longrightarrow\:\:\rm(\tan(A)-\sin(A))^2=(n)^2\\\\\\\longrightarrow\:\:\rm\tan^2(A)+\sin^2(A)-2\tan(A)\sin(A)=n^2\qquad\sf...eq(ll)$

$\rule{150}{1.5}$

✩ Subtracting eq. ( II) from eq. ( I) :

⠀

$:\implies\sf \tan^2(A)+\sin^2(A)+2\tan(A)\sin(A)=m^2\\\\:\implies\sf-\:(\tan^2(A)+\sin^2(A)-2\tan(A)\sin(A))=-\:n^2\\\dfrac{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}{}\\:\implies\sf 2\tan(A)\sin(A) + 2\tan(A)\sin(A) = m^2 - n^2\\\\\\:\implies\sf 4\tan(A)\sin(A)=m^2-n^2\\\\\\:\implies\sf 4\sqrt{\tan^2(A)\sin^2(A)}=m^2-n^2\\\\\\:\implies\sf 4\sqrt{\tan^2(A)(1-\cos^2(A))}=m^2-n^2\\\\\\:\implies\sf 4\sqrt{\tan^2(A)-\tan^2(A)\cos^2(A)}=m^2-n^2\\\\\\:\implies\sf 4\sqrt{\tan^2(A)-\sin^2(A)}=m^2-n^2\\\\\\:\implies\sf 4\sqrt{(\tan(A)+\sin(A))(\tan(A)-\sin(A))}=m^2-n^2\\\\\\:\implies\sf 4\sqrt{mn}=m^2-n^2\\\\\\:\implies\underline{\boxed{\sf m^2-n^2=4\sqrt{mn}}}$