Question

6. If secA + tanA = m and secA - tanA = n, find the value of
$\sqrt{mn}$
Standard:- 10

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Answer 1

Heya !!!



SecA + TanA = M


And,


SecA - TanA = N




Therefore,



✓MN = ✓ ( SecA + TanA ) ( SecA - TanA )



=> ✓ ( Sec²A ) - ( Tan²A )



We know that,


Sec theta = 1/Cos theta


And,


Tan theta = Sin theta / Cos theta


So,


✓ ( Sec²A ) - ( Tan²A )



=> ✓1/Cos² A - Sin²A / Cos²A



=> ✓ ( 1 - Sin²A ) / Cos²A



=> ✓Cos²A/Cos²A



=> ✓1 = 1



Hence,



✓MN = 1



★ HOPE IT WILL HELP YOU ★

Answer 2

$= > secA + tanA = m \: \: \: ...[Eq _{1}] \\ \\ AND \\ \\ = > secA - tanA = n \: \: \: ...[Eq _{2}]$
________________________________

●MULTIPLY BY Eq(1) AND Eq(2) , WE GET

$= > (secA - tanA) . (secA + tanA) = m \times n$
___________________________________

● USING ALGEBRAIC IDENTITY :-

$= >( \alpha + \beta ).( \alpha - \beta ) = ({ \alpha }^{2} - { \beta }^{2} )$
___________________________________

◢THEN

$= > (secA - tanA).(secA + tanA) = mn \\ \\ = > {sec}^{2} A - {tan}^{2} A = mn$
___________________________________

●USING TRIGONOMETRIC IDENTITY :-

$= > {sec}^{2} \alpha = 1 + {tan}^{2} \alpha \\ \\ = > {sec}^{2} \alpha - {tan}^{2} \alpha = 1$
____________________________________

◢SO,

$= > {sec}^{2} A - {tan}^{2} A = mn \\ \\ = > 1 = mn \\ \\ = > mn = 1$
__________________[◢ANSWER]

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