Question

5. ABC is an isosceles right-angled triangle, right angled at A, then
(a) sec B = cosec A
(b) sin B = cos C
(c) tan B = cot A
(d) cos B = sec B​

Answer 1

Answer:

sinB = cosA

May it helps you

Answer 2

b) sin B = cos C

Step-by-step explanation:

Given that

$\angle A = 90^\circ$ and $\triangle ABC$ is an isosceles triangle. so side AB =side AC

$\therefore \angle B = \angle C$       [ since AB = AC]

⇔$\angle B = \angle C = 45^\circ$    [∵ $\angle A +\angle B +\angle C = 180 ^ \circ$   ⇔   $90^\circ+\angle B +\angle B= 180^\circ$

                                                                           ⇔$\angle B= 45^\circ$]

a) sec B = cosec A

$sec 45^\circ$≠ $cosec 90^\circ$

b) sin B = cos C

$sin 45^\circ$ = $cos 45^\circ$

c) tan B = cot A

$tan 45 ^\circ$≠$cot 90^\circ$

d) cos B = sec B

$cos 45^\circ$≠$sec 45 ^ \circ$