Let us consider a right triangle ABC, right-angled at point B Given: As, a trigonometric Ratio shows the ratio between different sides, cot also shows the ratio of Base and Perpendicular. As we don't know the absolute value of Base and Perpendicular, let the base and perpendicular be multiplied by a common term, let it be kTherefore,Base = 7 kPerpendicular = 8kAccording to pythagoras theorem,(Hypotenuse)2 = (Base)2 + (Perpendicular)2 Applying Pythagoras theorem in ΔABC, we obtain AC2 = AB2 + BC2 = (8k)2 + (7k)2 = 64k2 + 49k2 = 113k2 AC = √113 k Now we have all the three sides of the triangle, Base = 7 kPerpendicular = 8 kHypotenuse = √113 kNow applying other trigonometric angle formulasSin = = = Cos θ = = = (i) Putting the obtained trigonometric ratios into the expression we get, = (1 – sin2 θ)/(1 – cos2 θ) = 49/64 (ii) Cot2 θ = (cot θ)2

Q. 74.0( 323 Votes )

If evaluate:

(i)

(ii)

Class 10th NCERT - Mathematics 8. Introduction to Trigonometry

Answer :

Let us consider a right triangle ABC, right-angled at point B

Given:

As, a trigonometric Ratio shows the ratio between different sides, cot also shows the ratio of Base and Perpendicular. As we don't know the absolute value of Base and Perpendicular, let the base and perpendicular be multiplied by a common term, let it be k
Therefore,
Base = 7 k
Perpendicular = 8k
According to pythagoras theorem,
(Hypotenuse)² = (Base)²+ (Perpendicular)²

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC = √113 k

Now we have all the three sides of the triangle,
Base = 7 k
Perpendicular = 8 k
Hypotenuse = √113 k
Now applying other trigonometric angle formulas