For all values of - the given expression is equal to- sin-cosec -2-cos-sec-2-tan2-cot2-

The correct option is D 7 (sinα+cosec nbsp;α)^2+(cosα+α)^2 (tan^2α+^2α) =sin^2α+cosec^2α+2+cos^2α+^2α+2 tan^2α ^2α =4+sin^2α+cos^2α+^2α t ...

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Trigonometric Identities

For all value...

Question

For all values of $α$ the given expression is equal to:
$(s i n α + c o s e c α)^{2} + (c o s α + s e c α)^{2} - (t a n^{2} α + c o t^{2} α)$

0

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$(s i n α + c o s e c α)^{2} + (c o s α + s e c α)^{2} - (t a n^{2} α + c o t^{2} α)$ for all values of $α$ is equal to

(sin α + cosec α)^{2} + (cos α + sec α)^{2} = k^{2} + {tan}^{2} α + {cot}^{2} α

k^{2} =

(sin α + c o s e c α)^{2} + (cos α + sec α)^{2} = k + {tan}^{2} α + {cot}^{2} α

cos (α + β) = \frac{4}{5}

sin (α - β) = \frac{5}{13}

0 \leq α, β \leq \frac{π}{4}

tan 2 α

cos (α + β) = \frac{3}{5}, sin (α - β) = \frac{5}{13}

0 < α, β < \frac{π}{4},

tan (2 α)

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