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प्रश्न
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
उत्तर
sin 3A = 1
sin 3A = sin90°
3A = 90°
A = 30°
`tan^2A – (1)/(cos^2"A") = tan^2 30° – (1)/(cos^2 30°)`
= `(1/sqrt3)^2 – (1)/(sqrt3/2)^2`
= `(1)/(3) – (4)/(3)`
= `(–3)/(3)`
= – 1
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