In a right-angled triangle, the sides are a, b and c with c as hypoten

1.	In a right-angled triangle, the sides are a, b and c with c as hypotenuse and c – b ≠ 1, c + b ≠ 1. Then the value of \(\bigg[\frac{log_{c+b}\,a+log_{c-b}\,a}{2\,log_{c+b}\,a\times\,log_{c-b}\,a}\bigg]\) is(a) –1 (b) \(\frac{1}{2}\)(c) 1 (d) 2
Answer» (c) 1 In a right angled triangle with a, b as sides and c as hypotenuse, c² = a² + b² (Pythagoras’ Theorem) Now, given expression = \(\frac{log_{c+b}\,a+log_{c-b}\,a}{2\times\,log_{c+b}\,a\times\,log_{c-b}\,a}\) = \(\frac{1}{2}\bigg[\frac{1}{\text{log}_{c-b}\,a}+\frac{1}{\text{log}_{c+b}\,a}\bigg]\) = \(\frac{1}{2}\big[\text{log}_a(c-b)+\text{log}_a(c+b)\big]\) = \(\frac{1}{2}\big[\text{log}_a(c-b)(c+b)\big]\) = \(\frac{1}{2}\big[\text{log}_a(c^2-b^2)\big]\) = \(\frac{1}{2}\text{log}_a\,a^2\) = log_a a = 1.

In a right-angled triangle, the sides are a, b and c with c as hypotenuse and c – b ≠ 1, c + b ≠ 1. Then the value of \(\bigg[\frac{log_{c+b}\,a+log_{c-b}\,a}{2\,log_{c+b}\,a\times\,log_{c-b}\,a}\bigg]\) is(a) –1 (b) \(\frac{1}{2}\)(c) 1 (d) 2

Answer»

(c) 1

In a right angled triangle with a, b as sides and c as hypotenuse,

c² = a² + b² (Pythagoras’ Theorem)

Now, given expression = \(\frac{log_{c+b}\,a+log_{c-b}\,a}{2\times\,log_{c+b}\,a\times\,log_{c-b}\,a}\)

= \(\frac{1}{2}\bigg[\frac{1}{\text{log}_{c-b}\,a}+\frac{1}{\text{log}_{c+b}\,a}\bigg]\)

= \(\frac{1}{2}\big[\text{log}_a(c-b)+\text{log}_a(c+b)\big]\) = \(\frac{1}{2}\big[\text{log}_a(c-b)(c+b)\big]\)

= \(\frac{1}{2}\big[\text{log}_a(c^2-b^2)\big]\) = \(\frac{1}{2}\text{log}_a\,a^2\) = log_a a = 1.