Inventive NCERT Class 10 Trigonometry Problem Solving Techniques for Lightning Quick Solutions: Set 4
Learn inventive NCERT class 10 trigonometry problem solving techniques to solve selected and similar trigonometry problems in a flash.
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.
Approach is to always analyze the problem for finding useful patterns and adopt applicable methods and strategies to solve the problems in mind in a few tens of seconds.
We will provide explanations along with solutions
The inventive problem solving showcased in these two solutions should also be very useful in preparation of competitive tests such as SSC CGL, SSC CGL TIER II, SSC CHSL, SSC CPO and similar MCQ based tests.
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 level Trigonometry problems—Set 4
Problem 1.
Prove the identity,
$\displaystyle\frac{\cos \text{A} - \sin \text{A} +1}{\cos \text{A} +\sin \text{A} - 1} =\text{cosec } \text{A}+\text{cot } \text{A}$
Solution 1. Problem analysis, Primary objective analysis and first pattern identification
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 deciding on the primary objective is obvious,
We must eliminate the denominator.
As such there is no direct way of doing that. But we observe the first key pattern,
All three terms of the numerator and denominator are same, except for the signs.
We make our first conclusion—eliminating the denominator will somehow depend on this first key pattern. We don't know how at this point but, we know, it must.
Can we use $(a+b)(a-b)=a^2-b^2$ on the numerator denominator? Briefly we looked into the possibility and mentally going through just one step rejected it.
Then we recalled a general trigonometric expression simplifying strategy that we have formed through our experience in solving many identities of complex forms. Let's state it,
In general, use of $(\sec \theta \pm \tan \theta)$ or $(\text{cosec } \theta \pm \text{cot } \theta)$ simplifies an expression involving $\sin \theta$ and $\cos \theta$ much faster and in fewer number of steps, than using the more well-known relation $(sin^2 \theta+cos^2 \theta=1)$.
The reason of supremacy of $\sec$-$\tan$ and $\text{cosec}$-$\text{cot}$ over $\sin$-$\cos$ in solving more complex problems lies in the inverse relationships between their complementary additive subtractive expressions,
$\sec \theta-\tan \theta=\displaystyle\frac{1}{\sec \theta + \tan \theta}$, and
$\text{cosec } \theta - \text{cot } \theta=\displaystyle\frac{1}{\text{cosec } \theta+\text{cot } \theta}$.
Because of the high simplifying potential in complex trigonometric problem solving, we consider these two pairs as primary friendly trigonometric function pairs. We include $\sin$-$\cos$ also in this category, but as a secondary pair of less potential.
Solution 1: End state analysis confirmed our decision to transform in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$
Following the Inventive End state analysis approach, a powerful general problem solving strategy, as we examined the RHS, presence of $(\text{cosec } \text{A}+\text{cot } \text{A})$ helped us make up our mind only to use this primary friendly trigonometric function pair for quick simplification.
We have to convert the LHS in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$.
As expected, reaching the result from this point took less than 20 secs in only a few steps.
Let us show you the deductive steps.
Solution 1: Problem solving execution: Deductive steps
Accordingly, converting the LHS in terms of, $\text{cot } \text{A}$ and $\text{cosec } \text{A}$,
LHS$=\displaystyle\frac{\cos \text{A} - \sin \text{A} +1}{\cos \text{A} +\sin \text{A} - 1}$
$=\displaystyle\frac{\text{cot } \text{A}-1+\text{cosec } \text{A}}{\text{cot } \text{A}+1-\text{cosec } \text{A}}$, dividing numerator and denominator by $\sin \text{A}$,
$=\displaystyle\frac{\text{cosec } \text{A} + \text{cot } \text{A}-1}{1-(\text{cosec } \text{A}-\text{cot } \text{A})}$, rearranging the terms for ease of understanding,
$=\displaystyle\frac{\text{cosec } \text{A}+\text{cot } \text{A} - 1}{1- \displaystyle\frac{1}{\text{cosec } \text{A}+\text{cot } \text{A}}}$
$=(\text{cosec } \text{A}+\text{cot } \text{A})\times{\displaystyle\frac{\text{cosec } \text{A}+\text{cot } \text{A} - 1}{\text{cosec } \text{A}+\text{cot } \text{A} - 1}}$
$=\text{cosec } \text{A}+\text{cot } \text{A}$
Proved.
You may try other methods.
This is a problem where the solution in a few steps is not directly visible and you have to go deeper and use your experience and concept based trigonometric problem solving strategies.
We will show a second approach to reasoning that enables you to take the crucial decision to convert the LHS in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$ faster.
Reasoning by Inventive Working backwards technique
The thinking goes like this from the end to the beginning—
- If at the end of simplification, only the factor $(\text{cosec } \text{A}+\text{cot } \text{A})$ is left, any other factors present before this final stage must have been cancelled out leaving only this factor,
- To have the factor $(\text{cosec } \text{A}+\text{cot } \text{A})$ in the numerator, the most suitable way is to have its complementary expression $(\text{cosec } \text{A}-\text{cot } \text{A})$ in the denominator, at the stage just before the previous stage above. Realize, we are moving backwards.
- To have the expression, $(\text{cosec } \text{A}-\text{cot } \text{A})$ in the denominator then, the given expression must be transformed to $\text{cosec } \text{A}$ and $\text{cot } \text{A}$,
- Finally reaching the beginning, we examine the given expression and find that, yes, this is the way to go.
This powerful inventive problem solving technique of Working backwards can be considered as an offshoot of more general problem solving approach of End state analysis.
Problem 2
Prove the identity,
$\sqrt{\displaystyle\frac{1+\sin \text{A}}{1-\sin \text{A}}}=\sec \text{A}+ \tan \text{A}$
Solution 2: Problem analysis, primary objective analysis and key pattern identification
The primary objective is obvious,
We have to transform the terms or expressions under square root of LHS in the form of squares to make those free of the square root.
The key pattern that would achieve this objective is also clear,
Multiply the denominator and numerator by $(1+\sin \text{A})$.
The numerator, free of square root becomes then, $(1+\sin \text{A})$, and the denominator, $\cos \text{A}$.
Straightforward division of the numerator by the denominator results in, $\sec \text{A} + \tan \text{A}$, the RHS.
Proved.
Let us show the steps.
Solution 2: Problem solving steps
LHS$=\sqrt{\displaystyle\frac{1+\sin \text{A}}{1-\sin \text{A}}}$
$=\sqrt{\displaystyle\frac{1+\sin \text{A}}{1-\sin \text{A}}\times{\displaystyle\frac{1+\sin \text{A}}{1+\sin \text{A}}}}$
$=\sqrt{\displaystyle\frac{(1+\sin \text{A})^2}{1-\sin^2 \text{A}}}$
$=\displaystyle\frac{1+\sin \text{A}}{\cos \text{A}}$
$=\sec \text{A}+\tan \text{A}$
Proved.
This is a less than twenty seconds problem solvable in mind.
Lessons learned: Inventive Concepts and techniques for lightning quick problem solving
Math specific inventive problem solving techniques
Denominator elimination technique:
- A general concept and technique in simplifying complex Algebraic and Trigonometric expressions is to eliminate the denominator in the fraction terms in the expressions (usually in the LHS). The concept shows the most promising way forward and helps to think of possible ways to do it.
Friendly trigonometric function pairs concepts:
- The friendly trigonometric function pairs $\sec$-$\tan$ and $\text{cosec}$-$\text{cot}$ derive their power in simplifying complex trigonometric expressions from the inverse relationships between their complementary additive and subtractive expressions,
- $\sec \theta-\tan \theta=\displaystyle\frac{1}{\sec \theta + \tan \theta}$, and
- $\text{cosec } \theta - \text{cot } \theta=\displaystyle\frac{1}{\text{cosec } \theta+\text{cot } \theta}$.
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.
Working backwards technique:
- This also is an effective broad-based problem solving technique that is used in problems such as Puzzles, daily life problems like forming the preparation schedule the moment the exam date is announced, target based project management decisions and so on.
- A good example of how this technique is used effectively is here.
- And the wide scope of its application in solving problems 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.