Q. 74.0( 323 Votes )

If cottheta = 7/8evaluate:

(i) (1+sintegrate heta ) (1-sintegrate heta )/(1+costheta ) (1-costheta )^ there there eξ sts

(ii) cot^2theta

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)+ (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

Sin =


=


=



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


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