Question

$if \\ \sin\alpha + \cos\alpha =1 \\ \\ prove \: that \\ \sin\alpha - cos \alpha = ( + - )1$
​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If sin α + cos α = 1, Then, prove that ( sin α - cos α) = ± 1.

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

• sin α + cos α = 1 .......(1)

$\Large{\underline{\mathfrak{\bf{\pink{To\:prove}}}}}$

• sin α + cos α = ± 1

$\Large{\underline{\underline{\mathfrak{\bf{Prove}}}}}$

We know,

★ ( a + b)² = a² + b² + 2ab

Squaring both side of equ(1)

➩ ( sin α + cos α )² = 1²

➩ ( sin² α + cos² α + 2 sin α .cos α) = 1

But,

★ ( sin² α + cos² α ) = 1

➩ ( 1 + 2 sin α .cos α) = 1

➩ 2 sin α .cos α = 0

➩ sin α .cos α = 0 ........(2)

Now, calculate,

• sin α - cos α = ?

we know,

★ ( a - b ) = ±√[(a+b)²-4ab]

So,

➩ sin α - cos α = ±√[(sin α + cos α)² - 4sin α .cos α ]

Keep value by equ(1) and equ(2)

➩ sin α - cos α = ±√[(1)²-4.0]

➩ sin α - cos α = ± √[1-0]

➩ sin α - cos α = ± 1

That's proved.

________________

$\Large{\underline{\mathfrak{\bf{\pink{Important\:Identity}}}}}$

★ ( sin² α + cos² α ) = 1

★ 2 sin α .cos α = sin 2α

★ cos 2α = sin² α - cos² α

= 1 - 2sin² α

= 2 cos² α - 1

= (1-tan²α)/(1+tan²α)

Answer 2

Given:

$\bold{ \sin( a ) + \cos( a ) = 1}$

To Prove:

$\bold{ \sin(a) - \cos(a) = \:( + - )1 }$

Proof:

We know,

★ ( a + b)² = a² + b² + 2ab

Squaring both side of equ(1)

➩ ( sin α + cos α )² = 1²

➩ ( sin² α + cos² α + 2 sin α .cos α) = 1

But,

★ ( sin² α + cos² α ) = 1

➩ ( 1 + 2 sin α .cos α) = 1

➩ 2 sin α .cos α = 0

➩ sin α .cos α = 0 ........(2)

Now, calculate,

sin α - cos α = ?

we know,

★ ( a - b ) = ±√[(a+b)²-4ab]

So,

➩ sin α - cos α = ±√[(sin α + cos α)² - 4sin α .cos α ]

Keep value by equ(1) and equ(2)

➩ sin α - cos α = ±√[(1)²-4.0]

➩ sin α - cos α = ± √[1-0]

➩ sin α - cos α = ± 1