Advertisements
Advertisements
Question
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Advertisements
Solution
LHS = `(1 - tan^2 θ)/(cot^2 θ - 1)`
= `(1 - tan^2 θ)/(1/tan^2 θ - 1)`
= `((1 - tan^2 θ)/(1 - tan^2 θ)/tan^2 θ) `
= tan2 θ
= RHS
Hence proved.
RELATED QUESTIONS
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
`(sec^2 theta-1) cot ^2 theta=1`
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
Prove that cot2θ × sec2θ = cot2θ + 1.
