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Cos[tan-1 1/3 + tan-1 1/2] = ______

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प्रश्न

`cos[tan^-1  1/3 + tan^-1  1/2]` = ______

विकल्प

  • `1/sqrt2`

  • `sqrt3/2`

  • `1/2`

  • `pi/4`

MCQ
रिक्त स्थान भरें
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उत्तर

`cos[tan^-1  1/3 + tan^-1  1/2] = \underlinebb((1/sqrt2))`

Explanation:

Let consider, y = `cos[tan^-1  1/3 + tan^-1  1/2]`

∵ `tan^-1 a + tan^-1 b = tan^-1 ((a + b)/(1 - ab))`, if ab < 1

∴ a = `1/3`; b = `1/2`; ab = `1/6 < 1`;

y = `cos [tan^-1 ((1/3 + 1/2)/(1 - 1/3 xx 1/2))]`

Thus, y = `cos [tan^-1 (((3 + 2)/6)/((6 - 1)/6))]`

y = `cos[tan^-1  5/5]`

y = cos[tan−1(1)]

y = `cos(pi/4)`

y = cos(45°)

∴ y = `1/sqrt2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०८]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 1.18 | पृष्ठ १०८

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