NCERT solutions for class 10 maths Trigonometry set 3

Submitted by Atanu Chaudhuri on Tue, 17/07/2018 - 14:13

Adopting algebraic expression simplifying strategies gave us the solutions fast enough

ncert solutions class 10 maths trigonometry set 3

In this session, we will take up two selected NCERT Trigonometry problems which we will solve in a few steps. In the process, we will highlight the problem solving approach, in which useful patterns, corresponding methods or algebraic expression simplifying strategies have been identified and applied for quick solution of the problems in a few steps.

Approach is to always analyze the problem for finding useful patterns and adopt associated methods if any. In the absence of any visible method on the pattern, we adopt algebraic expression simplifying strategies.

Algebraic expression simplifying often plays an important role in trigonometric expression simplification.

Generally, these problems are solved in mind. Attempt in these solutions always is to reduce the number as well as complexity of the steps to save valuable time even when writing the steps.

We will provide explanations along with solutions.

Let us start solving now.

Solutions to NCERT class 10 level Trigonometry problems—Set 3

Problem 1.

Prove the identity,

$\displaystyle\frac{\tan \theta}{1- \text{cot } \theta}+\displaystyle\frac{\text{cot } \theta}{1-\tan \theta}=1+\sec \theta.\text{cosec } \theta$

Solution 1. Problem analysis and applying Variable reduction and Denominator combining strategies

The inviolable rule that we follow in proving identities is to,

Apply operations on the LHS only to transform it to the RHS, that is, we move in a single direction, $\text{LHS} \Rightarrow \text{RHS}$.

On initial problem analysis we observed the LHS to be essentially a single variable expression, as $\text{cot } \theta=\displaystyle\frac{1}{\tan \theta}$.

Wherever such a situation arises, first we transform the expression to a simplified form by absorbing the inverses to transform the expression to a single variable one. We decided to transform $\text{cot } \theta$ to $\tan \theta$ just because the latter is generally recalled first among the two.

This strategy conforms to the highly effective general algebraic expression simplifying strategy of Reduction of number of variables that we call Variable reduction technique and state it as,

Reduce the number of variables in an algebraic expression at the first opportunity.

It is always easier to manipulate an algebraic (or trigonometric) expression with a single variable, rather than more than one variable.

We could visualize first denominator to turn to, $(\tan \theta - 1)$ with its numerator $\tan^2 \theta$. The second denominator is transformed to $\tan \theta(1-\tan \theta)$ with numerator $1$.

The useful pattern we find in the two denominators is the common factor of $(1-\tan \theta)$. This is the key pattern for us in this problem. It aids the second algebraic expression simplifying strategy of Minimum term denominator combining,

In an algebraic expression involving fractions of expressions, combine the denominators in the simplest way possible with minimum number of terms.

A general rule we follow,

We do not combine two algebraic expressions in the denominators if their product results in more than two terms.

In this situation, combining is simple enough so that next steps could be visualized easily without any mental load.

In the LCM of the denominator, the other factor $\tan \theta$ being a single term expression, visualization of the numerator as $(1-\tan^3 \theta)$ was easy. Automatically expansion of the numerator into its standard two factor form leads to cancellation of $(1-\tan \theta)$, leaving $(1+\tan \theta+\tan^2 \theta)$ in the numerator and $\tan \theta$ in the denominator.

One more division by $\tan \theta$ transforms the numerator to $(1+\tan \theta + \text{cot } \theta)$. Next step is straightforward.

Note: We have mentioned the trigonometric expressions as algebraic expressions frequently. This is essentially true with the qualification that the trigonometric expressions have additional trigonometric function relations to use for further and faster simplification. When you can't use the trigonometric function relations easily, use of the algebraic expression simplifying strategies coupled with general problem solving techniques and methods play the key role in simplification.

Let us show you the steps in writing.

Solution 1: Problem solving execution

We have decided to convert $\text{cot }\theta$ to $\tan \theta$ that transform the LHS,

LHS$=\displaystyle\frac{\tan \theta}{1- \text{cot } \theta}+\displaystyle\frac{\text{cot } \theta}{1-\tan \theta}$

$=\displaystyle\frac{\tan^2 \theta}{\tan \theta -1}+\displaystyle\frac{1}{\tan \theta(1-\tan \theta)}$

$=\displaystyle\frac{1-tan^3 \theta}{\tan \theta(1-\tan \theta)}$

Now we will use the expanded form of $(1-x^3)=(1-x)(1+x+x^2)$.

Taking up from the last stage,

LHS$=\displaystyle\frac{(1-\tan \theta)(1+\tan \theta +\tan^2 \theta)}{\tan \theta(1-\tan \theta)}$

$=\displaystyle\frac{1+\tan \theta+\tan^2 \theta}{\tan \theta}$

$=1 + \tan \theta + \text{cot } \theta$, dividing the numerator by $\tan \theta$ and rearranging to show the pairing of $\tan \theta$ and $\text{cot } \theta$

$=1+ \displaystyle\frac{1+ \tan^2 \theta}{\tan \theta}$

$=1+\sec^2 \theta.\text{cot } \theta$

$=1+\sec \theta.\text{cosec } \theta$

Proved.

We could have proved the identity through by a second method expressing $\tan \theta$ and $\text{cot } \theta$ in terms of $\sin \theta$ and $\cos \theta$. Let us show you the steps.

Solution 1. Second method

Converting $\tan \theta$ and $\text{cot }\theta$ in terms of $\sin \theta$ and $\cos \theta$, the LHS is transformed to,

LHS$=\displaystyle\frac{\tan \theta}{1- \text{cot } \theta}+\displaystyle\frac{\text{cot } \theta}{1-\tan \theta}$

$=\displaystyle\frac{\sin^2 \theta}{\sin \theta\cos \theta- \cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin \theta\cos \theta-\sin^2 \theta}$

$=\displaystyle\frac{\sin^2 \theta}{\cos \theta(\sin \theta- \cos \theta)}+\displaystyle\frac{\cos^2 \theta}{\sin \theta(\cos \theta-\sin \theta)}$

$=\displaystyle\frac{\sin^3 \theta -\cos^3 \theta}{\sin \theta\cos \theta(\sin \theta-\cos \theta)}$

$=\displaystyle\frac{\sin^2 \theta + \sin \theta\cos \theta+\cos^2 \theta}{\sin \theta\cos \theta}$

$=\displaystyle\frac{1 + \sin \theta\cos \theta}{\sin \theta\cos \theta}$, using $\sin^2 \theta +\cos^2 \theta=1$

$=1+\sec \theta\text{cosec } \theta$

Proved.

Choose the method that suits you.

This is what we call many ways technique. If we solve a problem in many ways, we become aware in our mind of possible different approaches to the solution and in addition would have the opportunity to compare various aspects of the solutions.

Essentially solving a problem in many ways improves your problem solving abilities.

Problem 2

Prove the identity,

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

Solution 2: Problem solving execution

We apply again a general algebraic expression simplifying strategy,

Eliminate the denominator at the earliest.

This makes the problem simpler in form and so, easier to manipulate.

Eliminating the denominator,

LHS$=1+\cos \text{A}$

With a look on the RHS, it was easy to decide multiplication of numerator and denominator by $(1-\cos \text{A})$ as the next action,

LHS$=\displaystyle\frac{1-\cos^2 \text{A}}{1-\cos \text{A}}$

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

Proved.

We have used End state analysis approach which recommends comparison of the goal state (the RHS here) with the initial or intermediate stage to see how similar the two stage results are. This is a natural problem solving approach.

This is a less than twenty seconds problem solvable in mind.

In fact if you attempt to solve such problems in mind using strategies, techniques, patterns and methods, not only would your mental exploration ability will increase, generally you would be able to solve a problem in fewer steps and in a lesser time in writing form also.