\begin{array} { l } { \text { (9) Prove that: } } \\ { \text { (i) } \

1.	\begin{array} { l } { \text { (9) Prove that: } } \\ { \text { (i) } \frac { \cos 27 ^ { \circ } + \sin 27 ^ { \circ } } { \cos 27 ^ { \circ } - \sin 27 ^ { \circ } } = \tan 72 ^ { \circ } } \\ { \text { (ii) } \frac { \cos 11 ^ { \circ } - \sin 11 ^ { \circ } } { \cos 11 ^ { \circ } + \sin 11 ^ { \circ } } = \cot 56 ^ { \circ } } \\ { \text { (iii) } \frac { \cos 15 ^ { \circ } - \sin 15 ^ { \circ } } { \cos 15 ^ { \circ } + \sin 15 ^ { \circ } } = \frac { 1 } { \sqrt { 3 } } } \end{array}
Answer» 1)cos27 + sin27 / cos27 - sin27 = tan72 LHS : cos27 + sin27 / cos27 - sin27 dividing throughtout by cos27 we get , 1 + tan27 / 1 - tan27 * 1 tan45 + tan27 / 1 - tan27. tan45 [ tan45 = 1 ] which is of the form TAN ( A + B ) = TanA + TanB / 1 - TanA TanB Therefore, tan ( 45 +27 ) = tan 72 2)LHS= cos11+sin11/cos11-sin11 divide both numerator and denominator by cos11 = 1+tan11/1- 1tan11 since tan 45=1 now we wil subtitute this value of 1 here in numerator =tan45+tan11/1-tan45tan11 (since it is the formula for tan (a+b)so) =tan(45+11) =tan56

\begin{array} { l } { \text { (9) Prove that: } } \\ { \text { (i) } \frac { \cos 27 ^ { \circ } + \sin 27 ^ { \circ } } { \cos 27 ^ { \circ } - \sin 27 ^ { \circ } } = \tan 72 ^ { \circ } } \\ { \text { (ii) } \frac { \cos 11 ^ { \circ } - \sin 11 ^ { \circ } } { \cos 11 ^ { \circ } + \sin 11 ^ { \circ } } = \cot 56 ^ { \circ } } \\ { \text { (iii) } \frac { \cos 15 ^ { \circ } - \sin 15 ^ { \circ } } { \cos 15 ^ { \circ } + \sin 15 ^ { \circ } } = \frac { 1 } { \sqrt { 3 } } } \end{array}

Answer»

1)cos27 + sin27 / cos27 - sin27 = tan72

LHS : cos27 + sin27 / cos27 - sin27

dividing throughtout by cos27

we get ,

1 + tan27 / 1 - tan27 * 1

tan45 + tan27 / 1 - tan27. tan45 [ tan45 = 1 ]

which is of the form TAN ( A + B ) = TanA + TanB / 1 - TanA TanB

Therefore, tan ( 45 +27 ) = tan 72

2)LHS= cos11+sin11/cos11-sin11

divide both numerator and denominator by cos11

= 1+tan11/1- 1tan11

since tan 45=1 now we wil subtitute this value of 1 here in numerator

=tan45+tan11/1-tan45tan11 (since it is the formula for tan (a+b)so)

=tan(45+11)

=tan56