Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you verify the identity $\sec (A-B)=\dfrac{\sec A\sec B}{1+\tan A\tan B}$?

Last updated date: 19th Jun 2024
Total views: 373.8k
Views today: 7.73k
Verified
373.8k+ views
Hint: In this question we will take the left-hand side of the expression which is $\sec (A-B)$and then take its reciprocal trigonometric function which is $\cos$and then use the angle addition-subtraction formula to expand the expression and then simplify the fraction to get the required right-hand side hence verifying the expression.

We have the expression as $\sec (A-B)=\dfrac{\sec A\sec B}{1+\tan A\tan B}$.
On taking the left-hand side of the expression, we get:
$\Rightarrow \sec (A-B)$
Now we know that $\sec \theta =\dfrac{1}{\cos \theta }$ therefore, on using this formula on the expression, we get:
$\Rightarrow \dfrac{1}{\cos (A-B)}$
Now since the denominator is in the form of addition-subtraction of trigonometric angles, we will expand it. We know that $\cos (a-b)=\cos a\cos b+\sin a\sin b$ therefore, on using the formula, we get:
$\Rightarrow \dfrac{1}{\cos A\cos B+\sin A\sin B}$
Now from the question, we want the term $1+\tan A\tan B$ in the denominator so to get that we will multiply and divide the numerator and denominator by $\cos A\cos B$.
On multiplying and dividing the numerator with $\cos A\cos B$, we get:
$\Rightarrow \dfrac{1}{\dfrac{\cos A\cos B\left( \cos A\cos B+\sin A\sin B \right)}{\cos A\cos B}}$
Now on splitting the denominator, we get:
$\Rightarrow \dfrac{1}{\cos A\cos B\left( \dfrac{\cos A\cos B}{\cos A\cos B}+\dfrac{\sin A\sin B}{\cos A\cos B} \right)}$
On simplifying the terms in the numerator, we get:
$\Rightarrow \dfrac{1}{\cos A\cos B\left( 1+\dfrac{\sin A\sin B}{\cos A\cos B} \right)}$
Now we know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$ therefore, on substituting it in the denominator, we get:
$\Rightarrow \dfrac{1}{\cos A\cos B\left( 1+\tan A\tan B \right)}$
Now the fraction can be split and written as:
$\Rightarrow \dfrac{\dfrac{1}{\cos A\cos B}}{1+\tan A\tan B}$
Now we know than $\dfrac{1}{\cos \theta }=\sec \theta$ therefore, on substituting it in the expression, we get:
$\Rightarrow \dfrac{\sec A\sec B}{1+\tan A\tan B}$, which is the right-hand side, hence proved.

Note: It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken.
The various trigonometric identities and formulae should be remembered while doing these types of sums. the various Pythagorean identities should also be remembered while doing these type of questions
To simplify any given equation, it is good practice to convert all the identities into and for simplifying.
If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.