Question

if A, B, C, D are interior angles of the quadrilateral then prove that sec(A+C)/4= Cosec(B+D)/4.

Answer 1

Answer:

plzzz give me brainliest ans and plzzzz follow me

Answer 2

Trigonometry is a branch of mathematics that studies the  relationships between side lengths and angles of triangles.

Step-by-step explanation:

Given: ABCD is a quadrilateral

To prove : $\sec (\dfrac{A+C}{4})=\text {cosec} (\dfrac{B+D}{4})$

Now as we know sum of all the angles of a quadrilateral is 360°

Therefore

A+B+C+D= 360°

A+C=360°-(B+C)

Now

$\sec(\dfrac{A+C}{4} )$

Put the value of A+C we get

$\sec (\dfrac{360^\circ -(B+D)}{4} )= \sec (90^\circ-\dfrac{(B+D)}{4} )$

Now as we know

$\sec (90^\circ -\theta) = \text {cosec }\theta$

Therefore

$\sec (90^\circ-\dfrac{(B+D)}{4} )= \text {cosec }(\dfrac{(B+D)}{4})$

Hence , proved the required result

#Learn more

Show that Sin(B+C/2) = cos(A/2). if A , B , C are the angles of a triangle ABC.

brainly.in/question/3632927