Advertisements
Advertisements
Question
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Advertisements
Solution 1
...
Solution 2
LHS = `(1 + tan^2A) + (1 + 1/tan^2A)`
= `(1 + sin^2A/cos^2A) + (1 + 1/(sin^2A/cos^2A))`
= `((cos^2A + sin^2A)/(cos^2A)) + ((cos^2A + sin^2A)/(sin^2A))`
= `1/(1 - sin^2A) + 1/sin^2A` (∵ `cos^2A + sin^2A = 1`)
= `(sin^2A + 1 - sin^2A)/(sin^2A(1 - sin^2A)) = 1/(sin^2A - sin^4A)`
APPEARS IN
RELATED QUESTIONS
`(sec^2 theta-1) cot ^2 theta=1`
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
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.
Choose the correct alternative:
1 + tan2 θ = ?
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2cosecθ`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`
