Question

$(1)tan { - }^{1 \binom{ cosx }{1 - sinx } } \frac{3\pi}{2 } < x < \frac{\pi}{2}$
को सरलतम रूप में व्यक्त कीजिए|

please answer
​

Answer 1

Answer:

??i dont know

Step-by-step explanation:

i have no clue lop

Answer 2

.tan

−1

1−sinx

cosx

Calculate cos x & 1-sin x .

As cos2x=cos

2

x−sin

2

x.

cos2(

2

x

)=cos

2

2

x

−sin

2

2

x

cosx=cos

2

2

x

−sin

2

2

x

Similarly sinx=2sin

2

x

cos

2

x

Now

tan

−1

[

1−(2sin

2

x

cos

2

x

)

cos

2

2

x

−sin

2

2

x

]

tan

−1

[

cos

2

2

x

+sin

2

2

x

−2sin

2

x

cos

2

x

cos

2

2

x

−sin

2

2

x

]

tan

−1

[

(cos

2

x

−sin

2

x

)

2

(cos

2

x

+sin

2

x

)(cos

2

x

−sin

2

x

)

]

tan

−1

[

cos

2

x

−sin

2

x

cos

2

x

+sin

2

x

]

tan

−1

[

cos

2

x

−sin

2

x

/cos

2

x

cos

2

x

+sin

2

x

/cos

2

x

]

tan

−1

[

1−tan

2

x

1+tan

2

x

]

tan

−1

(

1−tan

4

π

tan

2

x

tan

4

π

+tan

2

x

)

=tan

−1

[tan(

4

π

+

2

x

)]

=

4

π

+

2

x

.