RRB NTPC · प्रश्न 48
\(\frac{{sin\ {{27}^0}cos\ {{63}^0}}}{{co{s^2}\ {{27}^0}}} - \frac{{sec\ {{27}^0}cosec\ {{63}^0}}}{{ta{n^2}\ {{45}^0}}}\) आहे:
उत्तर दाखवा आणि स्पष्टीकरण
उत्तर: (D) -1
स्पष्टीकरण
cos 63° = sin 27°, so the first term is sin²27°/cos²27° = tan²27°. cosec 63° = sec 27° and tan 45° = 1, so the second term is sec²27°. tan²27° − sec²27° = −1.