If A - B - 90 - and A - 2 - 3 - then find the value of cosec B -

Given A+B=90^∘ sec A=23 sec(90^∘ B)=23 cosecB=23 ...

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Cos(A+B)Cos(A-B)

If A + B = ...

Question

If $A + B = 90^{\circ}$ and $sec A = \frac{2}{3},$ then find the value of cosec $B$ .

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90^{\circ}

c o t A

\sqrt{3}

c o s e c^{2} A

c o s e c A = \frac{15}{7}

A + B = 90

sec B

A = 45^{\circ} a n d 30^{\circ},

\frac{2 s i n^{2} (90^{\circ} - B) + s e c (90^{\circ} - A)}{c o s e c (90^{\circ} - B) + t a n^{2} (3 B - A)}

cosec A = \frac{4}{3}

A + B = 90

sec B =

If $c o s e c (A + B) = \frac{2}{\sqrt{3}}$ , sec(A-B)= $\frac{2}{\sqrt{3}}$ ,
$0^{\circ} < A + B \leq 90^{\circ}$ ,
Find A and B.

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