Two Problems for NCERT solutions for class 10 maths Trigonometry: by Inventive Math Problem Solving Techniques
Two problems for NCERT solutions for class 10 maths Trigonometry are solved in a few steps using a few powerful quick math problem solving techniques.
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 initial 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
In this solution, a more direct approach of reducing number of variables from 4 to 2 is chosen by converting $\text{cosec}$ to $\sin$ and $\text{sec}$ to $\cos$. This is more general Variable Reduction Technique frequently applied in solving tricky algebraic problems and equally useful in simplifying trigonometric expressions as well.
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 key pattern and method didcovery
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 $\sin \text{A}$. This is key pattern and method discovery, practically the basis of all inventive problem solving.
In this case, as in most problems, the key pattern is hidden one layer deep. The problem solver is the see through that hiding layer and discover the key pattern, crucial to solve the problem in a flash.
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.
Lessons learned: Inventive Concepts and techniques for lightning quick problem solving
Math specific inventive problem solving techniques
Variable reduction technique:
- A general algebraic technique in simplifying complex Algebraic and Trigonometric expressions is to reduce the number of variables in an expression which immediately reduces the complexity. A good example of application of this fundamental concept you will find here.
General problem solving techniques applicable to problems in many areas
End state analysis approach:
- This concept and technique forms an exceptionally effective general technique for solving complex problems in diverse areas such as Algebra, Trigonometry, Matchstick puzzles, Tough riddles, Real life problems and even in business process simplification.
- The power of this technique can be realized best from the solution of a matchstick puzzle you will find here.
Key pattern and method discovery technique:
- This is the foundation of all problem solving, right from the way a baby learns speak a language to solving a very complex business problem.
- An advanced example of compound key pattern in solving a rather difficult algebra problem you will find here.
- The mystery of human mind in identifying the key pattern in an apparent meaningless jumble you will find here.
End note
The approach to the solution of a problem is more important than the solution itself. It is a new way of thinking, not just solution of a problem.
Go through all NCERT Solutions Class 10 Maths Trigonometry here for solving all such problems confidently.