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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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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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  1. 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°

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