Question

In an isosceles triangle ABC;

AB = AC = 10 cm and BC = 18 cm.

Find the value of :

(i) sin²B + cos² C

ii) tan²C - sec² B + 2
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Answer 1

Answer:

1. sin²B + cos²C

=(AB/BC)²+ (AC/BC)²

25/81

2.tan²C- sec²B+2

=(AB/AC)²-(BC/AC)²+2

100/100-324/100 +2

=24/100

Answer 2

Please refer to the Attachment