Question

Sec 70- cot 40 (angles in degrees)

Answer 1

Given : Sec 70- cot 40 (angles in degrees)

To find : value

Solution

Sec 70 - Cot 40

= 1/Cos70 - Cos40/Sin40

= 1/Sin20 - Cos40/Sin40

sin2x = 2sinxcosx

= 1/Sin20 - Cos40/2Sin20Cos20

=(2Cos20 - Cos40)/2Sin20Cos20

=(2Cos(60-40) - Cos40)/Sin40

Cos(a-b) = cosacosb+sinasinb

= (2(Cos60Cos40 + Sin60Sin40) - Cos40)/Sin40

= (2 ((1/2) Cos40 + Sin60 Sin40 - Cos40)/Sin40

=( Cos40 + 2Sin60Sin40 - Cos40)/Sin40

= (2Sin60Sin40)/Sin40

= 2SIn60

=

$\sqrt{3}$

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Answer 2

Answer:

Hola ☺️

Step-by-step explanation:

Ur answer! ✌️✌️✌️✅✅✅

Sec 70 - Cot 40

use Sec = 1/Cos

& Cot = Cos/Sin

= 1/Cos70 - Cos40/Sin40

Use Cosx = Sin(90 - x)

= 1/Sin20 - Cos40/Sin40

sin2x = 2sinxcosx

= 1/Sin20 - Cos40/2Sin20Cos20

=(2Cos20 - Cos40)/2Sin20Cos20

=(2Cos(60-40) - Cos40)/Sin40

Cos(a-b) = cosacosb+sinasinb

= (2(Cos60Cos40 + Sin60Sin40) - Cos40)/Sin40

= (2 ((1/2) Cos40 + Sin60 Sin40 - Cos40)/Sin40

=( Cos40 + 2Sin60Sin40 - Cos40)/Sin40

= (2Sin60Sin40)/Sin40

= 2Sin60

= 2 ( root 3)/2

= root 3

Sec 70 - Cot 40 = root 3

Thanks ❤️❤️