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
What is \(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\) equal to?
If 3sin θ + 5cos θ = 5, then the value of 5sin θ - 3cos θ is equal to:
If angle C of a triangle ABC is a right angle where a, b and c are the sides opposite to the angles A, B and C respectively then what is tan A + tan B equal to?
If \(\sin \left( {A - B} \right) = \frac{1}{2}\) and \(\cos \left( {A + B} \right) = \frac{1}{2}\) , where A > B > 0° and A + B is an acute angle, then the value of A is:
In the equation
\(\rm\cos^{-1} \dfrac{1-a^2}{1 + a^2} - \cos^{-1} \dfrac{1-b^2}{1 + b^2} = 2 tan^{-1} x\) value of x is
