Consider the following :1. sin−154+sin−153=2π2. tan−13+tan−11=−tan−1(2+3)Which of the above is/are correct?

Solution:1.sin−154+sin−153=sin−1(54×54+53×53)∵sin−1x+sin−1y=sin−1(x1−y2+y1−x2)=sin−1(2516+259)=sin−11=2π2.tan−13+tan−11=tan−11−33+1=−tan−13−13+1=−tan−1(2+3) ..... [By rationalization]Both 1 and 2 are correct.Hence, C is the correct option.

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Consider the following : 1. {}^{-1}{4}{5}+{}^{-1}{3}{5}={}{2}

Question

Consider the following :
1. $sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{3}{5} = \frac{π}{2}$
2. $tan^{- 1} 3 + tan^{- 1} 1 = - tan^{- 1} (2 + 3)$
Which of the above is/are correct?

1 only

2 only

Both 1 and 2

Neither 1 nor 2

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Elijah Discussed

Consider the following :
1.

sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{3}{5} = \frac{π}{2}

tan^{- 1} 3 + tan^{- 1} 1 = - tan^{- 1} (2 + 3)

Which of the above is/are correct?

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Text solutionVerified

Solution:

1. sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{3}{5} = sin^{- 1} (\frac{4}{5} \times \frac{4}{5} + \frac{3}{5} \times \frac{3}{5})

∵ sin^{- 1} x + sin^{- 1} y = sin^{- 1} (x 1 - y^{2} + y 1 - x^{2})

= sin^{- 1} (\frac{16}{25} + \frac{9}{25})

= sin^{- 1} 1 = \frac{π}{2}

2. tan^{- 1} 3 + tan^{- 1} 1 = tan^{- 1} \frac{3 + 1}{1 - 3}

= - tan^{- 1} \frac{3 + 1}{3 - 1}

= - tan^{- 1} (2 + 3)

..... [By rationalization]

Both 1 and 2 are correct.

Hence, C is the correct option.

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Question Text	Consider the following : 1. $sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{3}{5} = \frac{π}{2}$ 2. $tan^{- 1} 3 + tan^{- 1} 1 = - tan^{- 1} (2 + 3)$ Which of the above is/are correct?
Answer Type	Text solution:1
Upvotes	150

Consider the following :1. sin−154​+sin−153​=2π​2. tan−13​+tan−11=−tan−1(2+3​)Which of the above is/are correct?

Text solutionVerified

Practice questions from similar books

Consider the following :
1. $sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{3}{5} = \frac{π}{2}$
2. $tan^{- 1} 3 + tan^{- 1} 1 = - tan^{- 1} (2 + 3)$
Which of the above is/are correct?