Question

if cos theta = 4/3 then cos^2 theta- sin^2 theta =​

Answer 1

Step-by-step explanation:

Answer:- 1/5

SOLUTION:-

cot(theta) = 4/3 = base(b) / perpendicular (p)

So, base(b) = 4 & perpendicular (p) = 3

Hence ,

hypotenus(h) = $\sqrt{ {4}^{2} + {3}^{2} }$hypotenus(h)=

4

2

+3

2

= $\sqrt{16 + 9}$=

16+9

= $\sqrt{25}$=

25

h = 5

Now,

cos^2(theta) - sin^2(theta)

= b/h - p/h

= 4/5 - 3/5

= 1/5

Answer 2

Answer:

Given:

• if cos theta = 4/3 then cos^2 theta- sin^2 theta

Solution:

= cot (theta) = 4/3

= base(b) / perpendicular (p)

So, base(b) = 4 & perpendicular (p) = 3

Hence ,

$\sf{h = \sqrt{ {4}^{2} + {3}^{2} } }$

$\sf{ = \sqrt{16 + 9} }$

$\sf{ = \sqrt{25} }$

h = 5

Now,

= b/h - p/h

= b/h - p/h= 4/5 - 3/5

= b/h - p/h= 4/5 - 3/5= 1/5

Hence,

the answer is 1/5