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InterviewSolution
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निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `tan^(-1)(tan.(3pi)/(4))` |
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Answer» चूँकि `tan^(-1)` की मुख्य शाखा का परिसर `(-(pi)/(2),(pi)/(2))` होता है। `therefore" "tan^(-1)(tan.(3pi)/(4))ne(3pi)/(4)," "[because (3pi)/(4) cancel in (-(pi)/(2),(pi)/(2))]` अब , `tan^(-1)(tan.(3pi)/(4))` `=tan^(-1)[tanpi-((pi)/(4))]," "[because (3pi)/(4)=pi-(pi)/(4)]` `=tan^(-1)[-tan((pi)/(4))],` `" "[because tan(pi-theta)=-tan theta]` `=tan^(-1)[tan(-(pi)/(4))]," "[because tan(-theta)=-tan theta]` `=-(pi)/(4) in (-(pi)/(2),(pi)/(2))," "[because tan^(-1)(tan theta)=theta]` `therefore" "tan^(-1)(tan.(3pi)/(4))=-(pi)/(4).` |
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