Question
Question: If \( \left( {x + y} \right) = {45^ \circ } \) , then prove that: (a) \( \left( {1 + \tan x} \righ...
If (x+y)=45∘ , then prove that:
(a) (1+tanx)(1+tany)=2
(b) (cotx−1)(coty−1)=2
Solution
Hint : The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as tan(x+y)=1−tanxtanytanx+tany and cot(x+y)=cotx+cotycotxcoty−1 . 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∘ .
First, we have to find the value of (1+tanx)(1−tanx) .
So, we know that (x+y)=45∘ .
Taking tangent on both sides of the equation, we get,
⇒tan(x+y)=tan45∘
Now, using the compound angle formula for tangent tan(x+y)=1−tanxtanytanx+tany on the left side of the equation, we get,
⇒1−tanxtanytanx+tany=tan45∘
We also know that the value of tan45∘ is one. So, we get,
⇒1−tanxtanytanx+tany=1
Cross multiplying the terms, we get,
⇒tanx+tany=1−tanxtany
Shifting all the terms consisting of trigonometric functions to the left side of the equation, we get,
⇒tanx+tany+tanxtany=1
Now, adding one to both sides of the equation, we get,
⇒tanx+tanxtany+tany+1=2
Now, taking tanx common from the first two terms and tany common from the last two terms in the left side of the equation, we get,
⇒tanx(1+tany)+(1+tany)=2
Factoring the equation, we get,
⇒(1+tanx)(1+tany)=2
So, we get the value of (1+tanx)(1+tany) as 2 .
Hence, proved.
Now, we also have to prove that (cotx−1)(coty−1)=2 .
Now, we have, (x+y)=45∘
Taking cotangent on both sides of the equation, we get,
⇒cot(x+y)=cot45∘
Now, using the compound angle formula for cotangent cot(x+y)=cotx+cotycotxcoty−1 in the left side of the equation, we get,
⇒coty+cotxcotxcoty−1=cot45∘
We also know that the value of cot45∘ is one. So, we get,
⇒coty+cotxcotxcoty−1=1
Cross multiplying the terms, we get,
⇒cotxcoty−1=coty+cotx
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,
⇒cotxcoty−coty−cotx=1
Now, adding one to both sides of the equation,
⇒cotxcoty−coty−cotx+1=2
Now, taking coty common from the first two terms and negative sign common from the last two terms, we get,
⇒coty(cotx−1)−(cotx−1)=2
Factoring the equation, we get,
⇒(cotx−1)(coty−1)=2
So, we get the value of (cotx−1)(coty−1) 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.