Question

what is the value of (cosx + sinx ) / (cosx - sinx ) 1) tan(pie / 4+ x ) 2) tan ( pie / 4-x ) 3) tanx 4) cotx​

Answer 1

Answer:

Answer

=[

sinx

1

−sinx][

cosx

1

−cosx][

cosx

sinx

+

sinx

cosx

]

=[

sinx

1−sin

2

x

][

cosx

1−cos

2

x

][

cosxsinx

sin

2

x+cos

2

x

]

=

sinx

cos

2

x

.

cosx

sin

2

x

.

cosxsinx

1

=1

Answer 2

Answer:

1st is the right option

$\underline{\underline{ tan \bigg(x + \dfrac{\pi}{4}\bigg)}}$

Step-by-step explanation:

Question :

$\dfrac{cosx+sinx}{cosx-sinx}$

To Find :

The value of given expression

Solution :

Dividing the numerator and denominator by cosx in the given expression

$: \implies \dfrac{\bigg(\dfrac{cosx+sinx}{cosx} \bigg)}{\bigg(\dfrac{cosx-sinx}{cosx} \bigg)}$

$: \implies \dfrac{1 + tanx}{1 - tanx}$

We know that tan45° = 1

So,

$: \implies \dfrac{1 + tanx}{1 -(1) tanx}$

$: \implies \dfrac{tan45^{\circ} + tanx}{1 - tan45^{\circ}\times tanx}$

We know that,

$\boxed{tan (A + B ) = \dfrac{tanA + tanB}{1 - tanAtanB} }$

Here, A = 45 and B = x

So,

$: \implies \dfrac{tan45^{\circ} + tanx}{1 - tan45^{\circ}\times tanx} = tan(45 + x)$

we know that in radians 45 is also written as π/4

So,

$: \implies tan(x + 45^{\circ} ) = tan \bigg(x + \dfrac{\pi}{4}\bigg)$

Hence the final result is

$\therefore \underline{ \boxed { \mathbf{{\dfrac{cosx+sinx}{cosx-sinx} = tan \bigg(x + \dfrac{\pi}{4}\bigg) }}}}$

Hope it helps!!