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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Q. Nos. 16 to 20 are of very short answer type questions.
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Question : 16
Total: 36
Find the value of sin‌−1[sin‌(−‌
17π
8
)].
Solution:     👈: Video Solution
sin‌−1(sin‌x)=x;‌
−π
2
≤x≤‌
π
2

sin‌−1[sin‌(−‌
17π
8
)].
=sin‌−1[−sin‌(‌
17π
8
)]=sin‌−1[−sin‌(2π+‌
π
8
)].
=sin‌−1[−sin‌(‌
17π
8
)]=sin‌−1[−sin‌(2π+‌
π
8
)].
=sin‌−1[−sin‌(‌
π
8
)]=sin‌−1[sin‌(‌
−π
8
)]
=−‌
π
8
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