InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `cos^(-1)(cos.(13pi)/(6))` |
|
Answer» चूँकि `cos^(-1)` की मुख्य शाखा का परिसर `[0,pi]` होता है। `therefore" "cos^(-1)(cos.(13pi)/(6))ne(13pi)/(6)," "[because(13pi)/(6) cancel in [0,pi]]` अब, `cos^(-1)(cos.(13pi)/(6))` `=cos^(-1)[cos(2pi+(pi)/(6))]," "[because(13pi)/(6)=2pi+(pi)/(6)]` `=cos^(-1)(cos.(pi)/(6))," "[because cos(2pi+theta)=cos theta]` `=(pi)/(6)in [0,pi]," "[because cos^(-1)(cos theta)=theta]` `therefore cos^(-1)(cos.(13pi)/(6))=(pi)/(6).` |
|