Question

if tan A + sin a is equal to M and tan a minus sin a is equal to N show that n square minus n square is equal to 4 root m n​

Answer 1

Given :-

$\bullet \sf \ tanA+sinA = m\\\\\bullet \ tanA-sinA = n$

To Prove :-

$\sf m^2-n^2= 4\sqrt{mn}$

Solution :-

$\bullet \sf \ L.H.S = m^2-n^2$

  $\longrightarrow \sf (tanA+sinA)^2-(tanA-sinA)^2\\\\\longrightarrow ( tan^2A+sin^2A+2tanAsinA)-(tan^2A+sin^2A-2tanAsinA) \\\\\longrightarrow tan^2A+sin^2A+2tanAsinA-tan^2A-sin^2A+2tanAsinA\\\\\longrightarrow 2tanAsinA+2tanA+sinA \\\\\longrightarrow 4tanAsinA$

$\bullet \sf \ R.H.S = 4 \sqrt{mn}$

   $\longrightarrow \sf 4\sqrt{(tanA+sinA)(tanA-sinA)} \\\\\longrightarrow 4\sqrt{tan^2A-sin^2A} \\\\\longrightarrow 4\sqrt{\dfrac{sin^2A}{cos^2A}-sin^2A} \\\\\longrightarrow 4\sqrt{\dfrac{sin^2A-(sin^2Acos^2A)}{cos^2A}} \\\\\longrightarrow 4\sqrt{ \dfrac{sin^2A(1-cos^2A)}{cos^2A} }$

$\bullet\bf \ 1-cos^2A = sin^2A$

    $\longrightarrow \sf 4\sqrt{sin^2A\times \dfrac{sin^2A}{cos^2A}} \\\\\longrightarrow 4\sqrt{sin^2Atan^2A} \\\\\longrightarrow 4 sinAtanA\\\\\longrightarrow 4tanAsinA$

∴ L.H.S = R.H.S

Hence proved.

Answer 2

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: QuEsTiOn:- \: \star}}}}}$

If tan A + sin A is equal to M and tan a minus sin a is equal to N show that m square minus n square is equal to 4 root m n

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

$\Large{\underline{\underline{\bf{Given:-}}}}$

❥ TanA + sinA = m

❥ TanA - sinA = n

$\Large{\underline{\underline{\bf{Show that:-}}}}$

❥ m² - n² = 4√mn

Here we go ,

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

$\sf \ L.H.S = m^2-n^2$

  $\longrightarrow \sf (tanA+sinA)^2-(tanA-sinA)^2\\\\\longrightarrow ( tan^2A+sin^2A+2tanAsinA)-(tan^2A+sin^2A-2tanAsinA) \\\\\longrightarrow tan^2A+sin^2A+2tanAsinA-tan^2A-sin^2A+2tanAsinA\\\\\longrightarrow 2tanAsinA+2tanA+sinA \\\\\longrightarrow 4tanAsinA$

$\sf \ R.H.S = 4 \sqrt{mn}$

   $\longrightarrow \sf 4\sqrt{(tanA+sinA)(tanA-sinA)} \\\\\longrightarrow 4\sqrt{tan^2A-sin^2A} \\\\\longrightarrow 4\sqrt{\dfrac{sin^2A}{cos^2A}-sin^2A} \\\\\longrightarrow 4\sqrt{\dfrac{sin^2A-(sin^2Acos^2A)}{cos^2A}} \\\\\longrightarrow 4\sqrt{ \dfrac{sin^2A(1-cos^2A)}{cos^2A} }$

$\bf \ 1-cos^2A = sin^2A$

    $\longrightarrow \sf 4\sqrt{sin^2A\times \dfrac{sin^2A}{cos^2A}} \\\\\longrightarrow 4\sqrt{sin^2Atan^2A} \\\\\longrightarrow 4 sinAtanA\\\\\longrightarrow 4tanAsinA$

$\mathfrak{\huge{\purple{\underline{\underline{Hence}}}}}$

LHS = RHS

Verified !