Find the value of a2-a2-14-a2-a2-14-

Putting a^2=x and √(a^2 1)=y, we have (a^2+√(a^2 1))^4+(a^2 √(a^2 1))^4 =(x+y)^4+(x y)^4 =[4_C_0 x^4 + 4_C_1 x^3y+ 4_C_2x^2y^2+4_C_3xy^3+4_C_4y^4] +[4_C_0 ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

Find the valu...

Question

Find the value of $(a^{2} + \sqrt{a^{2} - 1})^{4} + (a^{2} - \sqrt{a^{2} - 1})^{4}$ .

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Solution

Putting $a^{2} = x$ and $\sqrt{a^{2} - 1} = y$ , we have

$(a^{2} + \sqrt{a^{2} - 1})^{4} + (a^{2} - \sqrt{a^{2} - 1})^{4}$

$= (x + y)^{4} + (x - y)^{4}$

$= [4_{C_{0}} x^{4} + 4_{C_{1}} x^{3} y + 4_{C_{2}} x^{2} y^{2} + 4_{C_{3}} x y^{3} + 4_{C_{4}} y^{4}] + [4_{C_{0}} x^{4} - 4_{C_{1}} x^{3} y + 4_{C_{2}} x^{2} y^{2} - 4_{C_{3}} x y^{3} + 4_{C_{4}} y^{4}]$

$= 2 [4_{C_{0}} x^{4} + 4_{C_{2}} x^{2} y^{2} + 4_{C_{4}} y^{4}]$

$= 2 [x^{4} + 6 x^{2} y^{2} + y^{4}]$

$= 2 [(a^{2})^{4} + (a^{2})^{2} (\sqrt{a^{2} - 1})^{2} + (\sqrt{a^{2} - 1})^{4}]$

$= 2 [a^{8} + 6 a^{4} (a^{2} - 1) + (a^{2} - 1)^{2}]$

$= 2 [a^{8} + 6 a^{6} - 6 a^{4} + a^{4} - 2 a^{2} + 1]$

$= 2 [a^{8} + 6 a^{6} - 5 a^{4} - 2 a^{2} + 1]$

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Find the value of (a2+√a2−1)4+(a2−√a2−1)4.

Find the value of $(a^{2} + \sqrt{a^{2} - 1})^{4} + (a^{2} - \sqrt{a^{2} - 1})^{4}$ .