Find the values of (sin245° + cos245°)/tan260°

Find the values of

(i) (sin245° + cos245°)/tan260°

(ii) (sin30° – sin90° + 2cos0°)/tan2 60°

(iii) 4/3 tan230° + sin260° – 3cos260° + 3/4 tan260° – 2tan245°

Answer :

Trigonometric Ratios of Standard Angles Class 9 ICSE ML Aggarwal img 1

Trigonometric Ratios of Standard Angles Class 9 ICSE ML Aggarwal img 2

(iii) 4/3 tan230° + sin260° – 3cos260° + ¾tan260° – 2tan245°

= 4/3(1/√3)2 + (√3/2)2 – 3(1/2)2 + ¾×(√3)2 – 2×12

= (4/3× 1/3) + ¾ – (3 × ¼) + (¾ ×3) – (2×1)

= 4/9 + ¾ – 3/4 + 9/4 – 2

= 4/9 + 9/4 – 2

Taking LCM

= (16 + 81 – 72)/36

= (97 – 72)/36

= 25/36

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