2-1-a2 then cosec -3-

The correct option is B (2+a^2)^32 cosec α + ^3αα = 1sinα+cos^2αsin^3α=sin^2α+cos^2αsin^3α=1sin^3α = cosec^3 α = (1+^2α)^32 = (1 + ...

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Trigonometric Identities

2α=1+a2 then ...

Question

${cot}^{2} α = 1 + a^{2}$ then $cosec α + {cot}^{3} α sec α =$ ?

(1 + a^{2})^{\frac{2}{3}}

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(2 + a^{2})^{\frac{2}{3}}

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(1 + a^{2})^{\frac{3}{2}}

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(2 + a^{2})^{\frac{3}{2}}

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a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - . . . . + a_{2 n - 1}^{2} - a_{2 n}^{2}

(a_{1}, a_{2}, . . . . . . ., a_{10})

| a_{1} | \leq 1

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + . . . . . . . . + a_{10}^{2} - a_{1} a_{2} - a_{2} a_{3} - a_{3} a_{4} - . . . . . . - a_{9} a_{10} - a_{10} a_{1} = 2

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}

a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + . . . . . + a_{2 n - 1}^{2} - a_{2 n}^{2} = \frac{n}{2 n - 1} (a_{1}^{2} - a_{2 n}^{2})

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 2 & 1 & 5 1 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

a_{11} A_{21} + a_{12} A_{22} + a_{13} A_{23} =

a_{1}, a_{2}, a_{3}

S = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} +

a_{2 k}^{2}

= \frac{k}{2 k - 1} (a_{1}^{2} - a_{2 k}^{2})

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