1). $\frac{{{a^2} + \sqrt {{a^2} - 1} }}{a}$2). \(\frac{{{a^2} + \sq

1.	1). \(\frac{{{a^2} + \sqrt {{a^2} - 1} }}{a}\)2). \(\frac{{{a^2} + \sqrt {1 - {a^2}} }}{a}\)3). \(\frac{{{a^2} - \sqrt {{a^2} - 1} }}{a}\)4). 1
Answer» Sin(90°– a) = cos a Given, cos (10° 7’ 32”) = a Thus, sin (79° 52’ 28”) = a cos²a + sin²a = 1 sin(10° 7’ 32”) = √(1 – a²) tan(10° 7’ 32”) = $(\FRAC{{\sqrt {1 - {a^2}} }}{a})$ sin (79° 52’ 28”) + tan (10° 7’ 32”) $(= a + \frac{{\sqrt {1 - {a^2}} }}{a}\;)$ $(= \frac{{{a^2} + \sqrt {1 - {a^2}} }}{a})$

1). $\frac{{{a^2} + \sqrt {{a^2} - 1} }}{a}$2). $\frac{{{a^2} + \sqrt {1 - {a^2}} }}{a}$3). $\frac{{{a^2} - \sqrt {{a^2} - 1} }}{a}$4). 1

Answer»

Sin(90°– a) = cos a

Given, cos (10° 7’ 32”) = a

Thus, sin (79° 52’ 28”) = a

cos²a + sin²a = 1

sin(10° 7’ 32”) = √(1 – a²)

tan(10° 7’ 32”) = $(\FRAC{{\sqrt {1 - {a^2}} }}{a})$

sin (79° 52’ 28”) + tan (10° 7’ 32”)

$(= a + \frac{{\sqrt {1 - {a^2}} }}{a}\;)$

$(= \frac{{{a^2} + \sqrt {1 - {a^2}} }}{a})$