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Question : 154 of 180
Marks:
+1,
-0
Solution:
We have,
secθ=m and
tan‌θ=n‌ Now, ‌‌‌[(m+n)+‌]=‌‌[( secθ+tan‌θ)+‌]=‌‌[‌| ( secθ+tan‌θ)2+1 |
| ( secθ+tan‌θ) |
]=‌‌[‌| sec2θ+tan2θ+2 secθ‌tan‌θ+1 |
| ( secθ+tan‌θ) |
]=‌‌[‌| sec2θ+ sec2θ−1+2 secθ+tan‌θ+1 |
| secθ+tan‌θ |
]=‌‌[‌| 2 sec‌2θ+2 secθ‌tan‌θ |
| secθ+tan‌θ |
]=‌‌[‌| 2 secθ( secθ+tan‌θ) |
| ( secθ+tan‌θ) |
]=‌‌(2 secθ)=2
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