Question

If angle A = 90° in rt ∆ABC, prove that sec² B - cot² C = 1
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Answer 1

Step-by-step explanation:

Given a right angled triangle, right angled at A

=> ∠B + ∠C = 90°

ie., ∠C =  90° - ∠B

consider, sec² B - cot² C

= sec² B - cot² (90° - B)

= sec² B - cot² (90° - B)

= sec² B - tan² B

= 1                               (as sec²θ - tan²θ =1)

=> sec² B - cot² C = 1

hence proved