NCERT solutions for class 10 maths Trigonometry set 6

Submitted by Atanu Chaudhuri on Sun, 22/07/2018 - 02:47

Problem solving approach gives us the solutions in a few steps

ncert solutions class 10 maths trigonometry set 6

In this session, we will solve again two selected NCERT Trigonometry problems of class 10 standard in a few steps.

In the process, we will highlight the problem solving approach, in which, by applying useful pattern identification, suitable trigonometric expression simplifying strategies and problem solving techniques, quick solution of the problems could be achieved in a few steps.

We always analyze the problem for finding useful patterns and apply suitable methods to solve the problem in mind. This approach reduces the number of steps and speeds up the solution process in written form also.

We will provide explanations along with solutions.

Let us solve the problems to show you what we mean. We urge you to try to solve the problems yourself before going through the solutions.

Solutions to NCERT class 10 level Trigonometry problems—Set 6

Problem 1.

Prove the identity,

$(\text{cosec } \text{A}-\sin \text{A})(\sec \text{A}-\cos \text{A})=\displaystyle\frac{1}{\tan \text{A}+\text{cot } \text{A}}$

Solution 1: Problem analysis and useful pattern identification

Identifying the pattern that the RHS has $\tan$ and $\text{cot}$ functions, as well as identifying the possibility of obtaining $(\text{cosec}^2 \text{A}-1)$ in the numerator of the first factor and $(\sec^2 \text{A}-1)$, in the numerator of the second factor, we decide to transform the $\sin$ and $\cos$ to their inverses in the LHS. Solution comes quickly.

We apply here the End state analysis approach of comparing the target RHS expression with the intial given expression in LHS to decide on the method to use.

Let us show you the deductive steps.

Solution 1: Problem solving execution: Deductive steps

The LHS,

LHS$=(\text{cosec } \text{A}-\sin \text{A})(\sec \text{A}-\cos \text{A})$

$=\left(\text{cosec } \text{A}-\displaystyle\frac{1}{\text{cosec } \text{A}}\right)\left(\sec \text{A}-\displaystyle\frac{1}{\sec \text{A}}\right)$, in keeping with our decision

$=\displaystyle\frac{(\text{cosec}^2 \text{A}-1)(\sec^2 \text{A}-1)}{\text{cosec } \text{A}.\sec \text{A}}$

$=\displaystyle\frac{\text{cot}^2 \text{A}.\tan^2 \text{A}}{\displaystyle\frac{1}{\sin \text{A}.\cos \text{A}}}$

$=\displaystyle\frac{1}{\displaystyle\frac{\sin^2 \text{A}+\cos^2 \text{A}}{\sin \text{A}.\cos \text{A}}}$

$=\displaystyle\frac{1}{\tan \text{A}+\text{cot } \text{A}}$

$=RHS$.

Proved.

Let us show you the second way to solve the problem by inverting $\text{cosec } \text{A}$ and $\sec \text{A}$.

Problem 1: Second solution

LHS$=(\text{cosec } \text{A}-\sin \text{A})(\sec \text{A}-\cos \text{A})$

$=\left(\displaystyle\frac{1}{\sin \text{A}}-\sin \text{A}\right)\left(\displaystyle\frac{1}{\cos \text{A}}-\cos \text{A}\right)$

$=\displaystyle\frac{(1-\sin^2 \text{A})(1-\cos^2 \text{A})}{\sin \text{A}.\cos \text{A}}$

$=\displaystyle\frac{\cos^2 \text{A}.\sin^2 \text{A}}{\sin \text{A}.\cos \text{A}}$

$=\sin \text{A}.\cos \text{A}$

$=\displaystyle\frac{1}{\displaystyle\frac{1}{\sin \text{A}.\cos \text{A}}}$, by double inversion,

$=\displaystyle\frac{1}{\displaystyle\frac{\sin^2 \text{A}+\cos^2 \text{A}}{\sin \text{A}.\cos \text{A}}}$

$=\displaystyle\frac{1}{\tan \text{A}+\text{cot } \text{A}}$

$=RHS$.

Proved.

Choose your method.

Problem 2.

Prove the identity,

$\left(\displaystyle\frac{1+\tan^2 \text{A}}{1 + \text{cot}^2 \text{A}}\right)=\left(\displaystyle\frac{1-\tan \text{A}}{1- \text{cot } \text{A}}\right)^2=\tan^2 \text{A}$

Solution 2: Problem analysis and problem definition

The given problem is equivalent to two problems of proving identities,

Problem 2.1.

Prove the identity,

$\left(\displaystyle\frac{1+\tan^2 \text{A}}{1 + \text{cot}^2 \text{A}}\right)=\tan^2 \text{A}$.

And,

Problem 2.2.

Prove the identity,

$\left(\displaystyle\frac{1-\tan \text{A}}{1- \text{cot } \text{A}}\right)^2=\tan^2 \text{A}$.

The first problem is rather simple. Replace the numerator by equivalent $\sec^2 \text{A}$ and the denominator by the second equivalent, $\text{cosec}^2 \text{A}$. On simplification, it turns out to be $\tan^2 \text{A}$.

Let us show you the steps.

Solution 2.1: Problem solving steps

$LHS=\left(\displaystyle\frac{1+\tan^2 \text{A}}{1 + \text{cot}^2 \text{A}}\right)$

$=\displaystyle\frac{\sec^2 \text{A}}{\text{cosec}^2 \text{A}}$

$=\displaystyle\frac{\sin^2 \text{A}}{\cos^2 \text{A}}$

$=\tan^2 \text{A}$

$=RHS$

Proved.

Let us take the second part of the problem

Problem 2.2.

Prove the identity,

$\left(\displaystyle\frac{1-\tan \text{A}}{1- \text{cot } \text{A}}\right)^2=\tan^2 \text{A}$.

Solution 2.2: Problem analysis and pattern identification

In this case we concentrate on the fraction expression inside the square, and identify the common expression in numerator and denominator, $(\sin \text{A}-\cos \text{A})$ when we express the $\tan \text{A}$ and $\text{cot } \text{A}$ in terms of $\sin \text{A}$ and $\sin \text{A}$.

The common factor cancels out leaving $(-\tan \text{A})$ inside the bracket to be squared. When squared, we get the RHS.

Let us show you the steps.

Solution 2.2. Deductive steps

The LHS,

$LHS=\left(\displaystyle\frac{1-\tan \text{A}}{1- \text{cot } \text{A}}\right)^2$

$=\left(\displaystyle\frac{1-\displaystyle\frac{\sin \text{A}}{\cos \text{A}}}{1- \displaystyle\frac{\cos \text{A}}{\sin \text{A}}}\right)^2$

$=\left(\displaystyle\frac{-\sin \text{A}(\sin \text{A}-\cos \text{A})}{\cos \text{A}(\sin \text{A}-\cos \text{A})}\right)^2$

$=(-\tan \text{A})^2$

$=\tan^2 \text{A}$

$=RHS$.

Proved.

Proving both identities required well-used trigonometric expression simplifying patterns.

End note

We consider the approach to the solution more important than the solution itself. It is a way of thinking, not just solution of a problem.

If we try to solve a problem in mind using patterns and methods, and continue this practice, gradually speed of identifying patterns and applying associated methods improve significantly. Then, most problems of this type can be solved in mind in a few steps, saving valuable time for written deduction also.