Advertisements
Advertisements
Question
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Solution
LHS = `tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ)`
= `(sin^2θ/cos^2θ)/(sin^2θ/cos^2θ - 1) + (1/sin^2θ)/(1/(cos^2θ) - 1/sin^2θ)`
= `sin^2θ/(sin^2θ - cos^2θ) + (1/sin^2θ)/((sin^2θ - cos^2θ)/(cos^2θ sin^2θ))`
= `sin^2θ/(sin^2θ - cos^2θ) + cos^2θ/(sin^2θ - cos^2θ)`
= `(sin^2θ + cos^2θ)/(sin^2θ - cos^2θ) = 1/(sin^2θ - cos^2θ)` (∵`sin^2θ + cos^2θ = 1`)
Notes
θ
APPEARS IN
RELATED QUESTIONS
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ
