Reduce the following into sin x/cosx :-
${tan}^{4} x$

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Solution

${tan}^{4} x = {({tan}^{2} x)}^{2} ({tan}^{2} x + 1 = {sec}^{2} x) = {({sec}^{2} x - 1)}^{2} = {sec}^{4} x - 2 {sec}^{2} x + 1 = \frac{1}{{cos}^{4} x} - \frac{2}{{cos}^{2} x} + \frac{1}{1} = \frac{{cos}^{4} x - 2 {cos}^{2} x + 1}{{cos}^{4} x}$

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Similar questions

{tan}^{4} x

cos x

l i m x \to 0 \frac{1 - c o s^{3} x + s i n^{3} x + ℓ n (1 + x^{3}) + ℓ n (1 + c o s x)}{x^{2} - 1 + 2 c o s^{2} x + t a n^{4} x + s i n^{3} x}

s i n x . t a n 4 x = c o s x

π

t a n 4 x = \frac{4 t a n x (1 - t a n^{2} x)}{1 - 6 t a n^{2} x + t a n^{4} x}

y = x [(c o s x + s i n x) (c o s x - s i n x) + s i n x] + \frac{1}{2 \sqrt{x}}

\frac{d x}{d y}

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