Question

If we have a trigonometric ratio \[\cot \theta =\dfrac{7}{8}\], evaluate
(i) \[\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}\].
(ii) \[{{\cot }^{2}}\theta \].
(A). (i) \[\dfrac{44}{57}\]
       (ii) \[\dfrac{21}{29}\]
(B). (i) \[\dfrac{23}{34}\]
       (ii) \[\dfrac{19}{36}\]
(C). (i) \[\dfrac{49}{64}\]
       (ii) \[\dfrac{49}{64}\]
(D). None of these.

Answer 1

Hint: When an equation is given in terms of sine, cosine, tangent. We must use any of the trigonometric identities to make the equation solvable. These are many interrelations between sine, cosine, tan, secant these are interrelations that are called identities. Whenever you see conditions such that \[\theta \in R\], that means inequality is true for all angles. So, directly think of identity which will make your work easy.

Complete step-by-step solution -
Equality with sine, cosine, or tangent in them is called a trigonometric equation. These are solved by some inter-relations known before-hand.
All the inter relations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us never to result.Given value in the question, can be written as:
\[\cot \theta =\dfrac{7}{8}\] - (1)
By basic trigonometry, we know that cot, cos, sin are related as:
\[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]
\[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]
By above two conditions, we can combine them and say:
\[\cot \theta =\dfrac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }\] - (2)
By substituting the equation (2) in equation (1), we get:
\[\Rightarrow \dfrac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }=\dfrac{7}{8}\]
By squaring on both sides and cross multiplying, we get:
\[\Rightarrow 64\left( 1-{{\sin }^{2}}\theta \right)=49{{\sin }^{2}}\theta \]
By adding \[64{{\sin }^{2}}\theta \] on both sides of above equation we get:
\[\Rightarrow 113{{\sin }^{2}}\theta =64\]
By dividing with 113 and applying square root on both sides we get:
\[\Rightarrow \sin \theta =\dfrac{8}{\sqrt{113}}\] - (3)
By using \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\], we can say that the value of \[\cos \theta \] as:
\[\Rightarrow \cos \theta =\dfrac{7}{\sqrt{113}}\] - (4)
The first expression which we need to find the value is,
\[\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}\]
By using the identity \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\] on above expression we get,
\[\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\]
By substituting the equations (3), (4) in above term we get:
\[\Rightarrow \dfrac{1-{{\left( \dfrac{8}{\sqrt{113}} \right)}^{2}}}{1-{{\left( \dfrac{7}{\sqrt{113}} \right)}^{2}}}\]
By simplifying the square and taking L.C.M we get it as:
\[\Rightarrow \dfrac{113-64}{113-49}\]
By simplifying the square and taking L.C.M, we get it as: \[\dfrac{49}{64}\].
Second expression which we need to solve this question: \[{{\cot }^{2}}\theta \].
By substituting equation (1) into this term, we get it as: \[{{\left( \dfrac{7}{8} \right)}^{2}}\].
By simplifying this square, we can write its value as \[\dfrac{49}{64}\].
Therefore we can conclude \[\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}\]= \[\dfrac{49}{64}\], \[{{\cot }^{2}}\theta =\dfrac{49}{64}\].
Option (d) is correct.

Note: Alternately you can substitute \[1-{{\sin }^{2}}\theta \] as \[{{\cos }^{2}}\theta \] & \[1-{{\cos }^{2}}\theta \] as \[{{\sin }^{2}}\theta \] in first expression you can turn the expression into \[\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\] which is nothing but \[{{\cot }^{2}}\theta \]. By value you can get it directly instead of finding \[\sin \theta ,\cos \theta \]. After substituting \[\sin \theta ,\cos \theta \] generally students substitute in reverse. So, they get answers in reversed for \[\dfrac{64}{49}\] but it is wrong. So, substitute the values carefully to avoid such mistakes.