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Question: If the value of the trigonometric ratio, i.e. \({\text{cot}}\theta {\text{ = }}\dfrac{{\text{7}}}{8}...

If the value of the trigonometric ratio, i.e. cotθ = 78{\text{cot}}\theta {\text{ = }}\dfrac{{\text{7}}}{8} then find the value of (1 + sinθ)(1 - sinθ)(1 + cosθ)(1cosθ)\dfrac{{\left( {{\text{1 + sin}}\theta } \right)\left( {{\text{1 - sin}}\theta } \right)}}{{\left( {{\text{1 + cos}}\theta } \right)\left( {1 - {\text{cos}}\theta } \right)}}

Explanation

Solution

Hint: In this question, we have been given the value of cot function, using that and a basic trigonometric identity which includes sine and cosine function, we simplify the equation into known trigonometric identity. Then whenever it is required we will use the following formula. i.e.
sin2θ + cos2θ=1  {\text{si}}{{\text{n}}^2}\theta {\text{ + co}}{{\text{s}}^2}\theta = 1 \\\ \\\

Complete step-by-step answer:
Given data
cotθ = 78{\text{cot}}\theta {\text{ = }}\dfrac{{\text{7}}}{8}
Now (1 + sinθ)(1 - sinθ)(1 + cosθ)(1cosθ)\dfrac{{\left( {{\text{1 + sin}}\theta } \right)\left( {{\text{1 - sin}}\theta } \right)}}{{\left( {{\text{1 + cos}}\theta } \right)\left( {1 - {\text{cos}}\theta } \right)}}= 1 + sinθ - sinθ - sin2θ1 + cosθ - cosθ - cos2θ\dfrac{{{\text{1 + sin}}\theta {\text{ - sin}}\theta {\text{ - si}}{{\text{n}}^2}\theta }}{{{\text{1 + cos}}\theta {\text{ - cos}}\theta {\text{ - co}}{{\text{s}}^2}\theta }}
1 - sin2θ1 - cos2θ - Equation 1\Rightarrow \dfrac{{{\text{1 - si}}{{\text{n}}^2}\theta }}{{{\text{1 - co}}{{\text{s}}^2}\theta }}{\text{ - Equation 1}}

As we know sin2θ + cos2θ=1   {\text{si}}{{\text{n}}^2}\theta {\text{ + co}}{{\text{s}}^2}\theta = 1 \\\ \\\
sin2θ=1cos2θ and  cos2θ=1sin2θ  \Rightarrow {\text{si}}{{\text{n}}^2}\theta = 1 - {\text{co}}{{\text{s}}^2}\theta {\text{ and}} \\\ {\text{ }}{\cos ^2}\theta = 1 - {\text{si}}{{\text{n}}^2}\theta \\\

Using this relation to solve Equation 1 we get
1 - sin2θ1 - cos2θ=cos2θsin2θ. cot2θ (as cotθ = cosθsinθ)  \Rightarrow \dfrac{{{\text{1 - si}}{{\text{n}}^2}\theta }}{{{\text{1 - co}}{{\text{s}}^2}\theta }} = \dfrac{{{\text{co}}{{\text{s}}^2}\theta }}{{{\text{si}}{{\text{n}}^2}\theta }}. \\\ \Rightarrow {\text{co}}{{\text{t}}^2}\theta {\text{ }}\left( {{\text{as cot}}\theta {\text{ = }}\dfrac{{{\text{cos}}\theta }}{{{\text{sin}}\theta }}} \right) \\\
Therefore, (1 + sinθ)(1 - sinθ)(1 + cosθ)(1cosθ)\dfrac{{\left( {{\text{1 + sin}}\theta } \right)\left( {{\text{1 - sin}}\theta } \right)}}{{\left( {{\text{1 + cos}}\theta } \right)\left( {1 - {\text{cos}}\theta } \right)}}=(78)2=4964{\left( {\dfrac{7}{8}} \right)^2} = \dfrac{{49}}{{64}}.

Note: In such types of questions analyze the equations and perform basic mathematical operations. Then use trigonometric identities such that the required part of the problem can be reduced into a known or given trigonometric ratio. Trigonometric identities come in very handy for tackling this kind of problem.