If A- sin 2 - 2 - co 2 - 1-2 - and B- cos 2 - -tan 2 - - 2 - 1-2 - then A-B is equal to

The correct option is D [ 1 1; 1 1 ] Given, #160; A=[ sin ^ 2 α #160; #160; amp; ^ 2 α #160; #160;; co ^ 2 α ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration of Piecewise Continuous Functions

If A=[ sin ...

Question

If $A = [\begin{matrix} {sin}^{2} α & {sec}^{2} α c o {sec}^{2} α & 1 / 2 \end{matrix}]$ and $B = [\begin{matrix} {cos}^{2} α & - {tan}^{2} α - {cot}^{2} α & 1 / 2 \end{matrix}]$ , then $A + B$ is equal to

[\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\begin{matrix} 1 & 1 1 & 1 \end{matrix}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\begin{matrix} - 1 & 0 0 & - 1 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

(\frac{1}{{sec}^{2} α - {cos}^{2} α} + \frac{1}{csc^{2} α - {sin}^{2} α}) \times {sin}^{2} α {cos}^{2} α = \frac{1 - {sin}^{2} α {cos}^{2} α}{2 + {sin}^{2} α {cos}^{2} α}

(\frac{1}{{sec}^{2} α - {cos}^{2} α} + \frac{1}{{csc}^{2} α - {sin}^{2} α}) {cos}^{2} α . {sin}^{2} α = \frac{1 - {cos}^{2} α . {sin}^{2} α}{2 + {cos}^{2} α . {sin}^{2} α}

α

z = (1)^{1 / 5},

(1 + α + α^{2} + α^{- 2} - α^{- 1})

1, α, α^{2}, . . . . ., α^{n - 1}

n^{t h}

(1 - α) (1 - α^{2}) . . . . . (1 - α^{n - 1})

α

α = \sqrt[5]{1},

2^{∣ ∣ 1 + α + α^{2} + α^{3} - α^{- 1} - α^{- 2} ∣ ∣}

Join BYJU'S Learning Program