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NCERT solutions for class 10 maths Trigonometry set 5

Problem solving approach gives us the solutions in a few steps

ncert solutions class 10 maths trigonometry set 5

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, time tested trigonometric and algebraic expression simplifying strategies and powerful 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 and strategies to solve the problem in mind.

We will provide explanations along with solutions.

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

Solutions to NCERT class 10 level Trigonometry problems—Set 5

Problem 1.

Prove the identity,

$\displaystyle\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta} =\tan \theta$

Solution 1. Problem analysis and useful pattern identification

On initial problem analysis, the lightly hidden pattern of $(1-2\sin^2 \theta)$ in the numerator and $(2cos^2 \theta - 1)$ in the denominator could be identified just by taking out the factors, $\sin \theta$ and $\cos \theta$.

We know these two can be further simplified by using trigonometric relation,

$\sin^2 \theta+cos^2 \theta=1$.

Going ahead, the 1's in the two expressions are substituted by $\sin^2 \theta+\cos^2 \theta$ and immediately both the expressions become same,

$\cos^2 \theta - \sin^2 \theta$

The two cancel out leaving $\tan \theta$ as the final result.

Let us show you the deductions.

Solution 1: Problem solving execution: Deductive steps

The LHS,

LHS$=\displaystyle\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta}$

$=\displaystyle\frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-1)}$

$=\displaystyle\frac{\sin \theta(\sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-\sin^2 \theta - \cos^2 \theta)}$, substituting $\sin^2 \theta+\cos^2 \theta$ for 1 in both numerator and denominator,

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

$=\tan \theta$

$=RHS$.

Proved.

This is a problem in which the useful pattern was lightly hidden and the problem could be solved easily and quickly in mind.

Two useful relations to remember,

$1-2\sin^2 \theta=2\cos^2 \theta - 1=\cos^2 \theta - \sin^2 \theta$.

Problem 2

Prove the identity,

$(\sin \text{A} + \text{cosec } \text{A})^2 +(\cos \text{A} + \sec \text{A})^2=7 + \tan^2 \text{A}+ \text{cot}^2 \text{A}$

Solution 2: Problem analysis, key pattern identification and strategic decision

The key pattern that could be identified in both the squared expressions is—the terms are inverses to each other. In such cases, when the sum of two such terms are squared the middle terms are simplified to just numeric 2. This is a powerful property of inverses that we always use. It is an important problem solving pattern and method.


If you are curious, read our article on, Principle of interaction of inverses for elegant problem solving.


So the decision has been to expand the squared expressions. Simplifying, we get the RHS in two steps.

Let us proceed with the deductions.

Solution 2: Problem solving steps

LHS$=(\sin \text{A} + \text{cosec } \text{A})^2 +(\cos \text{A} + \sec \text{A})^2$

$=\sin^2 \text{A} + \cos^2 \text{A} + 4 + \text{cosec}^2 \text{A} + \sec^2 \text{A}$, expanding and rearranging the terms,

$=1 + 4 + 2 + (\text{cosec}^2 \text{A} -1) + (\sec^2 \text{A}-1)$

$=7 + \tan^2 \text{A}+ \text{cot}^2 \text{A}$

$=RHS$

Proved.

This is also a problem easily solvable in mind which you should always try. This approach will improve your ability to solve problems in fewer number of steps and hence faster.

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.


Further reading materials on Trigonometry

NCERT solutions for class 10 maths

NCERT solutions for class 10 maths Ttrigonometry Set 6

NCERT solutions for class 10 maths Trigonometry Set 5

NCERT solutions for class 10 maths Trigonometry Set 4

NCERT solutions for class 10 maths Trigonometry Set 3

NCERT solutions for class 10 maths Trigonometry Set 2

NCERT solutions for class 10 maths Trigonometry Set 1

Class 10 Maths

How to solve school math problems in a few direct steps Trigonometry 5

How to solve school math problems in a few steps and many ways Trigonometry 4

How to solve school math problems in a few simple steps Trigonometry 3

How to solve school math problems in a few simple steps Trigonometry 2

How to solve school math problems in a few steps Trigonometry 1

Tutorials, question and answer sets for competitive exams valuable for school level

You may refer to a fair amount of concise tutorials and MCQ type question and answer sets created for competitive exams at the page containing list of links,

Suresolv Trigonometry

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NCERT Solutions for Class 10 Maths

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NCERT Solutions for Class 10 Maths on Real numbers part 3, HCF and LCM by factorization and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 2, Euclid’s division algorithms, HCF and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions

Chapter 3: Linear Equations

NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection

NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form

NCERT solutions for class 10 maths Chapter 3 Linear equations 5 Algebraic solution by Cross Multiplication

NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination

NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution

NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions

NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.

Chapter 4: Quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 4 Nature of roots of a quadratic equation

Chapter 6: Triangles

NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons

Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles

Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 2 Ratio values for selected angles

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities


Chapter 8: Introduction to Trigonometry

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1