Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $ \left( {x + y} \right) = {45^ \circ } $ , then prove that:
(a) $ \left( {1 + \tan x} \right)\left( {1 + \tan y} \right) = 2 $
(b) $ \left( {\cot x - 1} \right)\left( {\cot y - 1} \right) = 2 $

seo-qna
Last updated date: 24th Jul 2024
Total views: 351.9k
Views today: 9.51k
Answer
VerifiedVerified
351.9k+ views
Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $ \tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ and $ \cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot x + \cot y}} $ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer:
In the given problem, we have to simplify and find the value of the given trigonometric expressions using compound angle formulae of tangent and cotangent.
Now, we are given the sum of two angles, x and y as $ {45^ \circ } $ .
First, we have to find the value of $ \left( {1 + \tan x} \right)\left( {1 - \tan x} \right) $ .
So, we know that $ \left( {x + y} \right) = {45^ \circ } $ .
Taking tangent on both sides of the equation, we get,
 $ \Rightarrow \tan \left( {x + y} \right) = \tan {45^ \circ } $
Now, using the compound angle formula for tangent $ \tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ on the left side of the equation, we get,
 $ \Rightarrow \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} = \tan {45^ \circ } $
We also know that the value of $ \tan {45^ \circ } $ is one. So, we get,
 $ \Rightarrow \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} = 1 $
Cross multiplying the terms, we get,
 $ \Rightarrow \tan x + \tan y = 1 - \tan x\tan y $
Shifting all the terms consisting of trigonometric functions to the left side of the equation, we get,
 $ \Rightarrow \tan x + \tan y + \tan x\tan y = 1 $
Now, adding one to both sides of the equation, we get,
 $ \Rightarrow \tan x + \tan x\tan y + \tan y + 1 = 2 $
Now, taking $ \tan x $ common from the first two terms and $ \tan y $ common from the last two terms in the left side of the equation, we get,
 $ \Rightarrow \tan x\left( {1 + \tan y} \right) + \left( {1 + \tan y} \right) = 2 $
Factoring the equation, we get,
\[ \Rightarrow \left( {1 + \tan x} \right)\left( {1 + \tan y} \right) = 2\]
So, we get the value of \[\left( {1 + \tan x} \right)\left( {1 + \tan y} \right)\] as $ 2 $ .
Hence, proved.
Now, we also have to prove that $ \left( {\cot x - 1} \right)\left( {\cot y - 1} \right) = 2 $ .
Now, we have, $ \left( {x + y} \right) = {45^ \circ } $
Taking cotangent on both sides of the equation, we get,
 $ \Rightarrow \cot \left( {x + y} \right) = \cot {45^ \circ } $
Now, using the compound angle formula for cotangent $ \cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot x + \cot y}} $ in the left side of the equation, we get,
 $ \Rightarrow \dfrac{{\cot x\cot y - 1}}{{\cot y + \cot x}} = \cot {45^ \circ } $
We also know that the value of $ \cot {45^ \circ } $ is one. So, we get,
 $ \Rightarrow \dfrac{{\cot x\cot y - 1}}{{\cot y + \cot x}} = 1 $
Cross multiplying the terms, we get,
 $ \Rightarrow \cot x\cot y - 1 = \cot y + \cot x $
Now, shifting all the terms consisting of trigonometric functions to the left side of the equation and all the constants to the right side of equation, we get,
 $ \Rightarrow \cot x\cot y - \cot y - \cot x = 1 $
Now, adding one to both sides of the equation,
 $ \Rightarrow \cot x\cot y - \cot y - \cot x + 1 = 2 $
Now, taking $ \cot y $ common from the first two terms and negative sign common from the last two terms, we get,
\[ \Rightarrow \cot y\left( {\cot x - 1} \right) - \left( {\cot x - 1} \right) = 2\]
Factoring the equation, we get,
\[ \Rightarrow \left( {\cot x - 1} \right)\left( {\cot y - 1} \right) = 2\]
So, we get the value of \[\left( {\cot x - 1} \right)\left( {\cot y - 1} \right)\] as \[2\].
Hence, proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.