(Cos 135° - Cos 120° )/(Cos 135+ Cos 120)=3-2√2
Answers
Answer:
Answer and Explanation:
To prove : \frac{\cos 135-\cos 120}{\cos 135+\cos 120}=3-2\sqrt2
cos135+cos120
cos135−cos120
=3−2
2
Proof :
Take LHS,
LHS=\frac{\cos 135-\cos 120}{\cos 135+\cos 120}LHS=
cos135+cos120
cos135−cos120
LHS=\frac{\cos (90+45)-\cos(90+30)}{\cos (90+45)+\cos (90+30)}LHS=
cos(90+45)+cos(90+30)
cos(90+45)−cos(90+30)
LHS=\frac{-\sin 45+\sin 30}{-\sin 45-\sin 30}LHS=
−sin45−sin30
−sin45+sin30
LHS=\frac{\sin 45-\sin 30}{\sin 45+\sin 30}LHS=
sin45+sin30
sin45−sin30
Substitute the trigonometric values,
LHS=\frac{\frac{1}{\sqrt2}-\frac{1}{2}}{\frac{1}{\sqrt2}+\frac{1}{2}}LHS=
2
1
+
2
1
2
1
−
2
1
LHS=\frac{\frac{2-\sqrt2}{2\sqrt2}}{\frac{2+\sqrt2}{2\sqrt2}}LHS=
2
2
2+
2
2
2
2−
2
LHS=\frac{2-\sqrt2}{2+\sqrt2}LHS=
2+
2
2−
2
Rationalize,
LHS=\frac{2-\sqrt2}{2+\sqrt2}\times \frac{2-\sqrt2}{2-\sqrt2}LHS=
2+
2
2−
2
×
2−
2
2−
2
LHS=\frac{(2-\sqrt2)^2}{2^2-(\sqrt2)^2}LHS=
2
2
−(
2
)
2
(2−
2
)
2
LHS=\frac{4+2-4\sqrt2}{4-2}LHS=
4−2
4+2−4
2
LHS=\frac{6-4\sqrt2}{2}LHS=
2
6−4
2
LHS=3-2\sqrt2LHS=3−2
2
LHS=RHSLHS=RHS
Hence proved.
Answer:
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