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1.

Prove that : `(tan theta)/(sectheta+1)-(tantheta)/(1-sectheta)=2cosec theta`

Answer» L.H.S `=(tantheta)/(sectheta+1)-(tantheta)/(1-sectheta)=tan theta((1)/(sectheta+1)+(1)/(sectheta-1))`
`=tantheta[(sectheta-1+sectheta+1)/((sectheta-1)(sectheta+1))]= tantheta*(2sectheta)/(sec^(2)theta-1)`
`=(2tanthetasectheta)/(tan^(2)theta)=(2sectheta)/(tantheta)=2xx(1)/(costheta)xx(costheta)/(sintheta)`
`=(2)/(sintheta)=2"cosec"theta=R.H.S.`
Discussion
2.

`(tantheta+2)(2 tantheta+1) = 5 tantheta+ sec^(2)theta`

Answer» False,
LHS = `(tantheta+2)(2 tantheta+1)`
`=2tan^(2)theta+4 tantheta+tantheta+2` `[therefore sec^(2)theta-tan^(2)theta=1]`
`=2sec^(2)theta + 5 tantheta=RHS`
Discussion
3.

Prove that : `tantheta+tantheta+tan(90^(@)-theta)=sectheta*sec(90^(@)-theta)`

Answer» L.H.S. `=tantheta+tan(90^(@)-theta)`
`=tantheta+cottheta`
`=(sintheta)/(costheta)+(costheta)/(sintheta)=(sin^(2)theta+cos^(2)theta)/(costhetasintheta)`
`=(1)/(costheta*sintheta)=sectheta*"cosec "theta`
`=sectheta*sec(90^(@)-theta)=R.H.S.`
Discussion
4.

If `1+sin^(2)theta = 3sinthetacostheta`, then prove that `tantheta=1` or `1/2`.

Answer» Given, `1+sin^(2)theta= 3sintheta.costheta`
On dividing by `sin^(2)theta` on both sides, we get
`1/(sin^(2)theta)+1=3.cottheta` `[therefore cottheta=(costheta)/(sintheta)]`
`rArr "cosec"^(2)theta + 1=3.cottheta` `["cosec"theta = 1/(sintheta)]`
`rArr 1+cot^(2)theta+1=3.cottheta` `["cosec"theta= 1/(sintheta)]`
`rArr 1+ cot^(2)theta+1 = 3.cottheta` `[therefore "cosec"^(2)theta-cot^(2)theta=1]`
`rArr cot^(2)theta-3cottheta+2=0`
`rArr cot^(2)theta - 2cottheta-cottheta+2=0` [by splitting the middle term]
`rArr cottheta(cottheta-2)-1(cottheta-2)=0`
`rArr (cottheta-2)(cottheta-1)=0 rArr cottheta=1` or 2
`rArr tantheta=1 or 1/2` `[therefore tantheta=1/(cottheta)]` Hence Proved.
Discussion
5.

The angleof elevation of the top of a tower from a certain point is `30o`. If theobserver moves 20 m towards the tower, the angle of elevation of the top ofthe tower increases by `15o`. Theheight of the tower is(a) 17.3m (b) 21.9 m (c) 27.3 m (d) 30 m

Answer» Let the height of the tower be h.
also, SR=x m, `anglePSR = theta`
Given that, QS = 20m
and `anglePQR = 30^(@)`
Now, in `DeltaPSR`,
`tantheta=(PR)/(SR) = h/x`
`rArr tantheta=h/x`
`rArr x=h/(tantheta)`................(i)
Now, in `DeltaPQR`,
`tan30^(@)=(PR)/(QR) = (PR)/(QS+SR)`
`rArr 20+x = h/(tan 30^(@)) = h/(1//sqrt(3))`
`rArr 20+x = hsqrt(3)`
`rArr 20+h/(tantheta)=hsqrt(3)` ..............(ii) From Eq. (i)
Since, after moving 20 m towards the tower the angle of elevation of the top increases by `15^(@)`.
i.e., `anglePSR = theta= anglePQR + 15^(@)`
`rArr theta=30^(@) + 15 = 45^(@)`
`therefore 20 + (h/tan 45^(@)) = hsqrt(3)`
`rArr 20 + h/1= hsqrt(3)`
`rAr 20 = hsqrt(3)-h`
`rArr h(sqrt(3)-1)=20`
`therefore h=(20)/(sqrt(3)-1).(sqrt(3)+1)/(sqrt(3)+1)` [by rationalisation]
`rAr = (20(sqrt(3)+1))/(3-1) = (20(sqrt(3)+1))/(2)`
`rArr =10(sqrt(3)+1)`m
Hence, the required height lof tower is `10(sqrt(3)+1)`m.
Discussion
6.

The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the tower window. At certain instant the angles of elevation of a balloon from these windows are observed to be `60^(@)` and `30^(@)`, respectively. Find the height of the balloon above the ground.

Answer» Let the height of the balloon from above the grond is H.
A and OP = `w_(2)R=w_(1)Q=x`
Given that, height of lower window from the above the ground `=w_(2)P=2m=OR`
Height of upper window from above in the lower window `=w_(1)w_(2)=4m=QR`
`therefore BQ=OB=(QR+RO)`
`=H-(4+2)`
`=H-6`
and `angleBw_(1)Q=30^(@)`
`rArr angleBw_(2)R=60^(@)`
Now, in `DeltaBw_(2)R`
`tan60^(@)= (BR)/(w_(2)R) = (BQ+QR)/(x)`
`rArr sqrt(3) = ((H-6)+4)/(x)`
`rArr x=(H-2)/(sqrt(3))`................(i)
and in `DeltaBw_(1)Q`,
`tan30^(@)= (BQ)/(w_(1)Q)`
`tan30^(@)= (H-6)/x = 1/sqrt(3)`
`rArr x=sqrt(3)(H-6)`...........(ii)
From Eqs. (i) and (ii), ltbgt `sqrt(3)(H-6) = ((H-2))/(sqrt(3))`
`3(H-6) = H-2 = 3H-18 = H-2`
`rArr 2H=16 rArr H=8`
So, the required height is 8m.
Hence, the required height of the ballon from above ground is 8 m.
Discussion