Question

if cosecx-cotx=1/3 then find value of cos²x-sin²x​

Answer 1

Answer:

7/25

Step-by-step explanation:

cosec x -cot x=1/3……….(i)

Hence (1/sin x )-(cos x/sin x)=1/3

Or (1-cos x)/sin x=1/3

3(1-cos x)=sinc

Squaring both gives

9(1–2cos x+cos^2 x)=sin^2 x

i.e. 9–18cos x+9cos^2 x=1-cos^2 x

Or 10cos^2 x-18 cos x+8=0

Or 5 cos^2 -9cos x+4=0

Hence cos x = [-(-9) ±√{(-9)^2–4*5*4}]/(2*5)

=[9±√{(81–80)}]/10=[9±1]/10 =1 or 4/5

cos x=1 doesn’t satisfy (i) Only cos x=4/5 satisfies (i)

Since cosx=4/5

hence sin x= √(1-cos^2 x)=√{1-(4/5)^2}=3/5

Hence cos^2 x-sin^2 x=(4/5)^2-(3/5)^2

=16/25–9/25=7/25

Answer 2

Answer:

7/25 this is the answer I hope it's helpful