Question

if m =cos theta-sin theta and n= cos theta+ sin theta. then the value of sec^2 theta is 4/(m+n)^2 can u tell me how is it?​

Answer 1

$\large\pink{❥Answer}$

n

m

+

m

n

=

mn

m+n

m+n=cosθ−sinθ+cosθ+sinθ=2cosθ

mn

=

(cosθ−sinθ)(cosθ+sinθ)

=

cos

2

θ−sin

2

θ

=

cos2θ

mn

m+n

=

cos2θ

2cosθ

=2

cos

2

θ−sin

2

θ

cos

2

θ

=2

cos

2

θ

cos

2

θ−sin

2

θ

1

n

m

+

m

n

=

1−tan

2

θ

2

Answer 2

Answer:

See the explanation

Step-by-step explanation:

Given $m=\cos\theta-\sin\theta$ and $n=\cos\theta+\sin\theta$

Add these two

$\Rightarrow m+n=\cos\theta-\sin\theta+\cos\theta+\sin\theta\\\Rightarrow m+n=2\cos\theta\\\Rightarrow \cos\theta=\frac{m+n}{2}\\\Rightarrow \sec\theta=\frac{1}{\cos\theta}=\frac{2}{m+n}\\\therefore \sec^2\theta=(\frac{2}{m+n})^2=\frac{2^2}{(m+n)^2}=\frac{4}{(m+n)^2}$

Hence proved