Question

Evaluate:
(i) $\dfrac{{{{\sin }^2}{{63}^ \circ } + {{\sin }^2}{{27}^ \circ }}}{{{{\cos }^2}{{17}^ \circ } + {{\cos }^2}{{73}^ \circ }}}$
(ii) $\sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ }$

Answer 1

1)
Hint:The expression containing trigonometric functions can be evaluated from the relation of trigonometric identities. This expression is related to trigonometric ratios of complementary angles, two angles are said to be complementary if their sums equals $90$ degrees. Then we apply the trigonometric identitiy to get the required answer.

Complete step-by-step answer:
Let,
$x = \dfrac{{{{\sin }^2}{{63}^ \circ } + {{\sin }^2}{{27}^ \circ }}}{{{{\cos }^2}{{17}^ \circ } + {{\cos }^2}{{73}^ \circ }}}$
We can perform the numerator first,
we know that, $63 + 27 = 90$
Since, $63 = 90 - 27$, we can substitute $63$ with $90 - 27$ .
So, numerator becomes,
 ${\sin ^2}(90 - 27) + {\sin ^2}{27^ \circ }$.
Similarly, we can perform the denominator,
we know $73 + 17 = 90$ .
Hence $17 = 90 - 73$ so we can substitute $17$ with $90 - 73$ .
So, the denominator becomes ${\cos ^2}(90 - 73) + {\cos ^2}{73^ \circ }$ .
Substitute the values in the above expression,
$x = \dfrac{{{{\sin }^2}(90 - 27) + {{\sin }^2}{{27}^ \circ }}}{{{{\cos }^2}(90 - 73) + {{\cos }^2}{{73}^ \circ }}}$
We know that,
$\cos (90 - \theta ) = \sin \theta $
$\sin (90 - \theta ) = \cos \theta $,
Using the above relation, the expression becomes,
\[x = \dfrac{{{{\cos }^2}{{27}^ \circ } + {{\sin }^2}{{27}^ \circ }}}{{{{\sin }^2}{{73}^ \circ } + {{\cos }^2}{{73}^ \circ }}}\]
By using the trigonometric identity, we know that,
${\sin ^2}A + {\cos ^2}A = 1$
Thus, by using the trigonometric identities the expression becomes,

Hence the value of the expression is,
\[
   \\
  \dfrac{{{{\cos }^2}{{27}^ \circ } + {{\sin }^2}{{27}^ \circ }}}{{{{\sin }^2}{{73}^ \circ } + {{\cos }^2}{{73}^ \circ }}} = 1 \\
   \\
 \]

Note:If an expression contains a trigonometric function firstly convert the trigonometric function into trigonometric identities. The relation of trigonometric identities in the expression is used to evaluate the value of the given expression.Students should remember the trigonometric identities and standard angles for solving these types of questions.

ii)
Hint:The expression containing trigonometric functions can be evaluated from the relation of trigonometric identities. This expression is related to trigonometric ratios of complementary angles, two angles are said to be complementary if their sums equals $90$ degrees.Applying the trigonometric identity we get the required answer.

Complete step-by-step answer:
Let,
$x = \sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ }$
Complete step-by-step solution:
$x = \sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ }$
As we know that,
 $65 + 25 = 90$
Hence $25 = 90 - 65$ so we can substitute $25$ with $90 - 25$ .
$
  x = \sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ } \\
  x = \sin {(90 - 65)^ \circ }\cos {65^ \circ } + \cos {(90 - 65)^ \circ }\sin {65^ \circ } \\
 $
We know that,
$\cos (90 - \theta ) = \sin \theta $
$\sin (90 - \theta ) = \cos \theta $
Applying the above relation in the expression to get,
$x = \cos {65^ \circ }\cos {65^ \circ } + \sin {65^ \circ }\sin {65^ \circ }$
We know $\sin x \times \sin x = {\sin ^2}x$ and $\cos x \times \cos x = {\cos ^2}x$
 then substitute in it.to get,
$x = {\cos ^2}{65^ \circ } + {\sin ^2}{65^ \circ }$
By using the identity,
${\sin ^2}A + {\cos ^2}A = 1$ ,
The above expression becomes,
$
  x = {\cos ^2}{65^ \circ } + {\sin ^2}{65^ \circ } \\
  x = 1 \\
 $
Hence,
The value of expression is,
 $\sin {25^ \circ }\cos {65^ \circ } + \cos {25^ \circ }\sin {65^ \circ } = 1$

Note:We can solve this type of question for that firstly we need to find an expression where it is related to trigonometric ratios of complementary angles, and then find the expression containing trigonometric functions can be evaluated from the relation of trigonometric identities.We can also solve this question by using formula $\sin(A+B)=\sin A \cos B + \cos A \sin B$ ,we get $A=25^\circ$ and $B=65^\circ$ So, $\sin (A+B) = \sin (65 + 25)=\sin 90 =1$ we get same answer.