Question

Prove that $ \sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right) = \left( {\sec \theta + \cos ec\theta } \right) $

Answer 1

Hint: In order to solve this question we will first solve L.H.S, we will solve in such a way that we will break all the terms in form of $ \sin \theta $ and $ \cos \theta $ in L.H.S and similarly we will break all the terms in $ \sin \theta $ and $ \cos \theta $ in R.H.S and the final answers of both sides will be compared if they are equal then this equation is proved while if unequal then unproved.

Complete step by step solution:
For solving this question we will first solve L.H.S so for solving this we will break all the terms in $ \sin \theta $ and $ \cos \theta $ :
 $ \sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right) $
Now on breaking the terms in $ \sin \theta $ and $ \cos \theta $ we will get:
 $ \sin \theta \left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }}} \right) + \cos \theta \left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }}} \right) $
Now we will take L.C.M inside the bracket:
 $ \sin \theta \left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta }}} \right) + \cos \theta \left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta }}} \right) $ ………………….(1)
Now on solving R.H.S:
 $ \sec \theta + \cos ec\theta $
Now on breaking in $ \sin \theta $ and $ \cos \theta $ we will get:
 $ \dfrac{1}{{\cos \theta }} + \dfrac{1}{{\sin \theta }} $
Now on further solving this we will get:
 $ \dfrac{{\sin \theta + \cos \theta }}{{\sin \theta \cos \theta }} $ ……………………(2)
We will put these two terms as stated in question:
 $ \sin \theta \left( {\dfrac{{\cos \theta + \sin \theta }}{{\cos \theta }}} \right) + \cos \theta \left( {\dfrac{{\sin \theta + \cos \theta }}{{\sin \theta }}} \right) = \dfrac{{\sin \theta + \cos \theta }}{{\sin \theta \cos \theta }} $
On cancelling all the possible terms from this equation we will get:
 $ \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} = \dfrac{1}{{\sin \theta \cos \theta }} $
Taking the L.C.M in L.H.S and on further solving this we will get:
 $ \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }} = \dfrac{1}{{\sin \theta \cos \theta }} $
So we get this at final as we know that the identity $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ so by applying this we will get:
 $ \dfrac{1}{{\sin \theta \cos \theta }} = \dfrac{1}{{\sin \theta \cos \theta }} $
So this is the final we get both the sides equal hence this equation is finally solved.

Note: In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others.