Advertisements
Advertisements
Question
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
Solution
L.H.S : (sinθ+cosecθ)2 +(cosθ+secθ)2
=sin2θ + cosec2θ + 2 +cos2θ + sec2θ + 2 `[because sin θ = 1/(cosecθ) " and cos "θ = 1/ (secθ)]`
= sin2θ + cos2θ+1+cot2θ+1+tan2θ+4 `[because cosec^2θ+1+cot^2θ" and "sec^2 θ =1+ tan^2θ]`
=sin2θ+cos2θ+1+cot2θ+1+tan2θ+4 `[because cosec^2θ+1+cot^2θ " and" sec^2 θ=1 +tan^2θ]`
=1+1+1+4+tan2θ+cot2θ `[because cos^2θ+ sin^2θ=1]`
=7+ tan2θ+cot2θ
L.H.S-R.H.S
APPEARS IN
RELATED QUESTIONS
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following trigonometric identities.
if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove that:
tan (55° + x) = cot (35° – x)
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec 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
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
If cos A + cos2A = 1, then sin2A + sin4 A = ?