Given that AB = 29 units & BC = 21 units and ∠ABC =θ°& ∠ACB = 90°.

Using Pythagoras theorem in right angled triangle ∆ACB,

AB2 = AC2 + BC2

⇒ AC2 = AB2 − BC2

⇒ AC =\(\sqrt{AB^2-BC^2}\)=\(\sqrt{29^2-21^2}\)=\(\sqrt{841-441}=\sqrt{400}\)

=20 units.

(∵ BC = 21, AB = 29)

Now, cosθ =\(\frac{BC}{AB}=\frac{21}{29}.\)(∵ BC = 21, AB = 29)

And sinθ =\(\frac{AC}{AB}=\frac{20}{29}.\)(∵ AC = 20 and AB = 29)

Now, cos2θ− sin2θ =\((\frac{21}{29})^2-(\frac{20}{29})^2\)=\(\frac{21^2-20^2}{29^2}\)

=\(\frac{441 \,- \,400}{841}=\frac{41}{841}.\).

Hence Proved