Given:
tanθ = 7/8
⇒ Assume in a right triangle:
height = 7, base = 8, hypotenuse = √113
sinθ = 7/√113, cosθ = 8/√113, cotθ = 8/7
Now the expression:
[(1 + sinθ)(1 − sinθ)] / [(1 + cosθ)(1 − cosθ)(cotθ)]
= (1 − sin²θ) / [(1 − cos²θ)(cotθ)]
= cos²θ / [sin²θ · cotθ]
= cos²θ / [sin²θ · (cosθ/sinθ)]
= cos²θ / (sinθ cosθ)
= cosθ / sinθ
= cotθ
= 8/7
Thus the answer = 8/7
Important Questions from Trigonometric Identities
If cos 2θ = sin θ and θ lies between 0 and 90°, then θ will be:
\(\frac{3 - 4\sin^2 \theta}{\cos^2 \theta} + 2\tan^2 \theta\) can be simplified as:
Simplify \( \left(\frac{1}{\sin^2 A} - 1\right) \), where \( 0 < A \leq 90^\circ \).
If \( \tan \theta = \frac{8}{15} \), then the value of \( \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \) is:
If the volume of a cuboid is \(3x^2 - 27\), then its possible dimensions are:
