Question

Anyone please answer!!​

Answer 1

Answer:

$sinA=\dfrac{m^2-1}{m^2+1}$

Step-by-step explanation:

Note : Theta is written as A.

Method 1

Given,

secA + tanA = m

On squaring both sides :

= > ( secA + tanA )^2 = m^2

From the properties of expansion :

• ( a + b )^2 = a^2 + b^2 + 2ab

Then,

= > sec^2 A + tan^2 A + 2secAtanA = m^2 ...( 1 )

Therefore,

= > $\dfrac{m^2 - 1 }{m^2 + 1}$

Substituting the value of m from ( 1 ):

$\implies\dfrac{sec^2 A + tan^2 A + 2secAtanA - 1}{ sec^2 A + tan^2 A + 2secAtanA + 1}\\\\\\\implies \dfrac{sec^2 A - 1 + tan^2 A + 2secAtanA}{sec^2 A + tan^2 +1 + 2secAtanA}\\\\\\\implies \dfrac{(sec^2 A - 1 ) + tan^2 A + 2secAtanA}{sec^2 A + (tan^2 +1) + 2secAtanA}$

From the properties of trigonometry :

• sec^2 ∅ - 1 = tan² ∅
• tan^2 ∅ + 1 = sec² ∅

$\implies \dfrac{tan^2 A + tan^2A + 2secAtanA}{ sec^2A + sec^2A + 2secAtanA}\\\\\\\implies \dfrac{2tan^2A+2secAtanA}{2sec^2A+2secAtanA}\\\\\\\implies \dfrac{2tanA( tanA+secA)}{2secA(secA+tanA)}\\\\\\\implies \dfrac{tanA}{secA}\\\\\\\implies \dfrac{\dfrac{sinA}{cosA}}{\dfrac{1}{cosA}}\\\\\\\implies sinA$

Hence,

sinA = sinA

Left Hand Side = Right Hand Side

Method 2

On squaring both sides :

= > ( secA + tanA )^2 = m^2

From the properties of expansion :

( a + b )^2 = a^2 + b^2 + 2ab

Then,

$\implies (sec A + tan A)^2=m^2$

By Using Componendo and Dividendo, from ratio and proportion :

$\implies\dfrac{(secA+tanA)^2+1}{(secA+tanA)^2 - 1}=\dfrac{m^2+1}{m^2-1}$

$\implies\dfrac{(secA+tanA)^2-1}{(secA+tanA)^2 +1}=\dfrac{m^2-1}{m^2+1}$

$\implies\dfrac{sec^2 A + tan^2 A + 2secAtanA - 1}{ sec^2 A + tan^2 A + 2secAtanA + 1}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{sec^2 A - 1 + tan^2 A + 2secAtanA}{sec^2 A + tan^2 +1 + 2secAtanA}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{(sec^2 A - 1 ) + tan^2 A + 2secAtanA}{sec^2 A + (tan^2 +1) + 2secAtanA}=\dfrac{m^2-1}{m^2+1}$

From the properties of trigonometry :

sec^2 ∅ - 1 = tan² ∅

tan^2 ∅ + 1 = sec² ∅

$\implies \dfrac{tan^2 A + tan^2A + 2secAtanA}{ sec^2A + sec^2A + 2secAtanA}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{2tan^2A+2secAtanA}{2sec^2A+2secAtanA}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{2tanA( tanA+secA)}{2secA(secA+tanA)}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{tanA}{secA}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies \dfrac{\dfrac{sinA}{cosA}}{\dfrac{1}{cosA}}=\dfrac{m^2-1}{m^2+1}\\\\\\\implies sinA=\dfrac{m^2-1}{m^2+1}$

Hence proved.

Answer 2

Here is your answer ☝

HOPE IT HELPS YOU ⭐

#ALLENITE ❤