Question

if tan theta + sin theta = m, tan theta - sin theta = n,
then show that m square - n square = 4 root mn

Answer 1

Heya !!!


Tan theta + Sin theta = M



And,


Tan theta - Sin theta = N




To prove :- M² - N² = 4✓MN



We have


LHS = ( M² - N² )


=> ( Tan theta + Sin theta )² - ( Tan theta - Sin theta )²


As we know that,

( A + B)² - (A - B)² = 4AB



So,


=> ( Tan theta + Sin theta)² - (Tan theta -Sin theta )² = 4Tan theta Sin theta



RHS = 4 ✓MN



=> 4✓ ( Tan theta + Sin theta )(Tan theta -Sin theta)


As we know that,


( A + B ) ( A - B) = ( A)² - (B)²



=> 4✓(Tan² theta - Sin² theta )




=> 4 ✓ (Sin² theta/ Cos² theta - Sin² theta )



=> 4 × ✓ ( Sin² theta - Sin² theta Cos² theta )/Cos² theta )



=> 4 × Sin theta × ✓1-Cos² theta / Cos theta



=> 4 Tan theta × ✓Sin² theta



=> 4Tan theta Sin theta



Thus,


LHS = RHS = 4Tan theta Sin theta




Hence,



(M² - N² ) = 4✓MN






★ HOPE IT WILL HELP YOU ★

Answer 2

$given \: \: that \: \: \tan( \alpha ) + \sin( \alpha ) = m \: \: \: and \: \: \: \: \: \tan( \alpha ) - \sin( \alpha ) = n \\ \\ \\ \\ \\ \\ \\ => \tan( \alpha ) = \frac{m + n}{2} \: \: \: \: \: and \: \: \: \: \: \sin( \alpha ) = \frac{m - n}{2} \\ \\ \\ => \cot( \alpha ) = \frac{2}{m + n} \: \: \: \: \: and \: \: \: \cosec( \alpha ) = \frac{2}{m - n} \\ \\ \\ \\ using \: \cosec ^{2} ( \alpha ) - \cot^{2} ( \alpha ) = 1 \\ \\ \\ ( \frac{2}{m - n} ) ^{2} - ( \frac{2}{m + n} )^{2} = 1 \\ \\ \\ => 4(( {m + n)}^{2} - {(m - n)}^{2} ) = {(m - n)}^{2} {(m + n)}^{2} \\ \\ \\ => 4(( {m}^{2} + {n}^{2} + 2mn - {m}^{2} - {n}^{2} + 2mn) = ( {m}^{2} - {n}^{2} ) ^{2} \\ \\ \\ => 4(4mn) = {( {m}^{2} - {n }^{2}) }^{2} \\ \\ \\ \sqrt{4 \times 4mn } = {m}^{2} - {n}^{2} \\ \\ \\ => 4 \sqrt{mn} = {m}^{2} - {n}^{2}$



Hence, proved.





I hope this will help you


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