How i solve the tricky question of introduction of trigonometry of ncert class 10 please help me to find the best solution of question number 7 of exercise 8.3 and give me the easiest solution of this question. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
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Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. Q.7
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Given:
sin 67° + cos 75°
In term of sin as cos function and cos as sin function, it can be written as follows
sin 67° = sin (90° – 23°)
cos 75° = cos (90° – 15°)
So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)
Now, simplify the above equation
= cos 23° + sin 15°
Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°