Question

The value of (sin^(6)u+cos^(6)u-1)/(sin^(4x+cos^(4x-1) (wherever defined) is equal to​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\dfrac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1}}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{\dfrac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1}}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Formula used:}}$

$\mathsf{(i)\;a^3+b^3=(a+b)^3-3ab(a+b)}$

$\mathsf{(ii)\;a^2+b^2=(a+b)^2-2ab}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1}}$

$\mathsf{=\dfrac{(sin^2x)^3+(cos^2x)^3-1}{(sin^2x)^2+(cos^2x)^2-1}}$

$\textsf{Using formula (i) and (ii)}$

$\mathsf{=\dfrac{(sin^2x+cos^2x)^3-3(sin^2x\,cos^2x)(sin^2x+cos^2x)-1}{(sin^2x+cos^2x)^2-2\,sin^2x\,cos^2x-1}}$

$\mathsf{using\;the\;identity}\;\;\boxed{\mathsf{sin^2A+cos^2A=1}}$

$\mathsf{=\dfrac{(1)^3-3(sin^2x\,cos^2x)(1)-1}{(1)^2-2\,sin^2x\,cos^2x-1}}$

$\mathsf{=\dfrac{1-3\,sin^2x\,cos^2x-1}{1-2\,sin^2x\,cos^2x-1}}$

$\mathsf{=\dfrac{-3\,sin^2x\,cos^2x}{-2\,sin^2x\,cos^2x}}$

$\mathsf{=\dfrac{3}{2}}$

$\implies\boxed{\mathsf{\dfrac{sin^6x+cos^6x-1}{sin^4x+cos^4x-1}=\dfrac{3}{2}}}$

$\underline{\textbf{Find more:}}$

If sin q + cos q = √ 3 , then prove that tan q + cot q = 1.​

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Answer 2

Answer:

Step-by-step explanation: