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If sin 3A = 1 and 0 < A < 90°, find tan 2 A − 1 cos 2 A - Mathematics

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Question

If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`

Sum

Solution

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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Trigonometric Equation Problem and Solution
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Chapter 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (C) [Page 298]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (C) | Q 6.3 | Page 298
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