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Prove that tan^(–1) 63/16 = sin^(–1) 5/13 + cos^(–1) 3/5.

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प्रश्न

Prove that `tan^(-1)  63/16 = sin^(-1)  5/13 + cos^(-1)  3/5`.

प्रमेय
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उत्तर

Let `sin^(-1)  5/13 = x`.

Then, `sin x = 5/13` ⇒ `cos x = 12/13`.

∴ `tan x = 5/12` ⇒ `x = tan^-1  5/12`

∴ `sin^-1  5/13 = tan^-1  5/12`   ...(1)

Let `cos^-1  3/5 = y`.

Then, `cos y = 3/5` ⇒ `sin y = 4/5`.

∴ `tan y = 4/3` ⇒ `y = tan^-1  4/3`

∴ `cos^-1  3/5 = tan^-1  4/3`   ...(2)

Using (1) and (2), we have

R.H.S. = `sin^-1  5/13 + cos^-1  3/5`

= `tan^-1  5/12 + tan^-1  4/3`

= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  ((15 + 48)/(36 - 20))`

= `tan^-1  63/16`

= L.H.S.

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अध्याय 2: Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 [पृष्ठ ३१]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Miscellaneous Exercise on Chapter 2 | Q 7. | पृष्ठ ३१

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