Advertisements
Advertisements
Question
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Advertisements
Solution
LHS = `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ `
= `cos θ/cos θ + cot θ/cot θ + sec θ/sec θ`
= 1 + 1 + 1
= 3
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Choose the correct alternative:
cos 45° = ?
