$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

  • A
    $2$
  • B
    $4$
  • C
    $1$
  • D
    $0$

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Similar Questions

$\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ} = \ldots$

सिद्ध कीजिए कि $(\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta = 2$.

निम्नलिखित सर्वसमिकाओं को सिद्ध कीजिए,जहाँ कोण न्यून कोण हैं जिनके लिए व्यंजक परिभाषित हैं:
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

Difficult
View Solution

$\sin (90^\circ - \theta) = \ldots \ldots \ldots$

$\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ} = \ldots \ldots \ldots \ldots$

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