Trigonometry (Basic)

Study Material (PDF)
Self Assesment
Study Material (PDF)
Self Assesment

Click on each question to see answer & solution.

Question No: #131

If `\angle A + \angle B = 90^\circ`, then show that, `1 + tan A/tan B = sec^2 A`.

Question No: #132

If `A + B = 90^\circ`, then show that, `tan A + tan B = \frac{\text{cosec}^2 B}{\sqrt{\text{cosec}^2 B - 1}}`.

Question No: #133

If the angles `\alpha` and `\beta` are complimentary to each other, then prove that, `cot \beta + cos \beta = \frac{cos \beta}{cos \alpha} (1 + sin \beta)`.

Question No: #134

If `\angle P + \angle Q = 90^\circ`, then show that, `\sqrt{\frac{sin P}{cos Q} - sin P cos Q} = cos P`.

Question No: #135

If the angles `\alpha` and `\beta` are complimentary to each other, then show that, `sin \alpha = \sqrt{\frac{cos \alpha}{sin \beta} - cos \alpha sin \beta}`.

Question No: #136

If `A + B = 90^\circ`, then show that, `\sqrt{\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{sin^2 B}{cos^2 A}} = tan A`.

Question No: #137

If `\alpha + \beta = \pi/2`, then prove that, `\frac{sec \alpha + sin \beta}{sin \alpha} = tan \alpha + 2 tan \beta`.

Question No: #138

If the angles `A` and `B` are complimentary to each other, then prove that, `(sin A + sin B)^2 = 1 + 2 sin A cos A`.

Question No: #139

If `\alpha + \beta = 90^\circ`, then show that, `sec^2 \alpha + sec^2 \beta = sec^2 \alpha \cdot sec^2 \beta`.

Question No: #140

If `sin (A+B) = 1` and `cos(A-B) = 1`, where `0^\circ le (A+B) \le 90^\circ`, then find the value of `A` and `B`.
Ans: `A = B = 45^\circ`
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