Question

Evaluate: \[\dfrac{{\sin {{25}^ \circ }}}{{\sec {{65}^ \circ }}} + \dfrac{{\cos {{25}^ \circ
}}}{{\cos {\text{ec}}{{65}^ \circ }}}\]

Answer 1

Hint:We are given with an expression and we are asked to evaluate this expression. First try to simplify the given expression, recall the trigonometric identities and use them to simplify the given expression. Use the trigonometric identities for the required terms one by one.

Complete step by step solution:
Given, the expression \[P = \dfrac{{\sin {{25}^ \circ }}}{{\sec {{65}^ \circ }}} + \dfrac{{\cos {{25}^ \circ
}}}{{\cos {\text{ec}}{{65}^ \circ }}}\] (i)

Now, let us simplify the expression further.
We will use some trigonometric identities for simplification.

Secant is the inverse of cosine, so \[\sec \theta \] can be written as \[\dfrac{1}{{\cos \theta }}\]. Therefore, we can write \[\sec {65^ \circ }\] as \[\dfrac{1}{{\cos {{65}^ \circ }}}\]. Substituting this value in equation (i), we get

\[P = \dfrac{{\sin {{25}^ \circ }}}{{\dfrac{1}{{\cos {{65}^ \circ }}}}} + \dfrac{{\cos {{25}^ \circ }}}{{\cos
{\text{ec}}{{65}^ \circ }}}\]
\[ \Rightarrow P = \sin {25^ \circ }\cos {65^ \circ } + \dfrac{{\cos {{25}^ \circ }}}{{\cos {\text{ec}}{{65}^
\circ }}}\] (ii)

Cosecant is inverse of sine, so \[{\text{co}}\sec \theta \] can be written as \[\dfrac{1}{{\sin \theta }}\].

Therefore, we can write \[{\text{co}}\sec {65^ \circ }\] as \[\dfrac{1}{{\sin {{65}^ \circ }}}\].
Substituting this value in equation (ii) we get,
\[P = \sin {25^ \circ }\cos {65^ \circ } + \dfrac{{\cos {{25}^ \circ }}}{{\dfrac{1}{{\sin {{65}^ \circ }}}}}\]
\[ \Rightarrow P = \sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ }\] (iii)

We can write \[{65^ \circ }\] as \[{90^ \circ } - {25^ \circ }\], substituting this in equation (iii),

we get
\[P = \sin {25^ \circ }\cos \left( {{{90}^ \circ } - {{25}^ \circ }} \right) + \cos {25^ \circ }\sin \left( {{{90}^
\circ } - {{25}^ \circ }} \right)\] (iv)
\[\cos \left( {{{90}^ \circ } - \theta } \right)\] can be written as \[\sin \theta \]. So, \[\cos \left( {{{90}^
\circ } - {{25}^ \circ }} \right)\] can be written as \[\sin {25^ \circ }\]. Therefore equation (iv) becomes
\[P = \sin {25^ \circ }\sin {25^ \circ } + \cos {25^ \circ }\sin \left( {{{90}^ \circ } - {{25}^ \circ }} \right)\]

\[ \Rightarrow P = {\sin ^2}{25^ \circ } + \cos {25^ \circ }\sin \left( {{{90}^ \circ } - {{25}^ \circ }}
\right)\] (v)

Similarly, \[\sin \left( {{{90}^ \circ } - \theta } \right)\] can also be written as \[\cos \theta \].

So, \[\sin
\left( {{{90}^ \circ } - {{25}^ \circ }} \right)\] can be written as \[\cos {25^ \circ }\]. Therefore, equation (v) becomes,

\[P = {\sin ^2}{25^ \circ } + \cos {25^ \circ }\cos {25^ \circ }\]

\[ \Rightarrow P = {\sin ^2}{25^ \circ } + {\cos ^2}{25^ \circ }\] (vi)

We have the trigonometric identity, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] so using this identity for

\[{\sin ^2}{25^ \circ } + {\cos ^2}{25^ \circ }\] we have,
\[{\sin ^2}{25^ \circ } + {\cos ^2}{25^ \circ } = 1\]
Substituting this value in equation (vi), we get
\[P = 1\]

Therefore, \[\dfrac{{\sin {{25}^ \circ }}}{{\sec {{65}^ \circ }}} + \dfrac{{\cos {{25}^ \circ }}}{{\cos
{\text{ec}}{{65}^ \circ }}} = 1\]

Note:In trigonometry, there are three important functions, these are sine, cosine and tangent. There are three more trigonometric functions which can be written in terms of sine, cosine and tangent. These are cosecant, secant and cotangent. Cosecant is inverse of sine, that is \[\cos {\text{e}}c\theta {\text{ =
}}\dfrac{1}{{\sin \theta }}\]. Secant is inverse of cosine, that is \[\sec \theta {\text{ = }}\dfrac{1}{{\cos
\theta }}\] and cotangent is inverse of tangent, that is \[\cot \theta {\text{ = }}\dfrac{1}{{\tan \theta
}}\].