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Prove that: 3 cosec2 60°  - 2 cot2 30°  + sec2 45°  = 0 - Mathematics

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प्रश्न

Prove that:

3 cosec2 60°  - 2 cot2 30°  + sec2 45°  = 0

योग

उत्तर

LHS =3 cosec260° – 2 cot230° + sec245°

=`3(2/sqrt3)^2 – 2(sqrt3)^2 + (sqrt2)^2`

= `3xx4/3-2xx3+2`

= 4 – 6 + 2

= 0

= RHS

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अध्याय 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (A) [पृष्ठ २९१]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (A) | Q 3.6 | पृष्ठ २९१

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