Question

If `3secalpha-5tanalpha=k` and `6secalpha+ktanalpha=5` then what is the value of `k^2` among the following option?

`a)20`
`b)25`
`c)34`
`d)37`

Answer 1

Answer is (a) as `k^2=20`

Explanation:

We have `3secalpha-5tanalpha=k` .......(A) and

`6secalpha+ktanalpha=5` .......(B) and

Multiplying (A) by `2` and subtracting from (A), we get

`(k+10)tanalpha=5-2k` and `tanalpha=(5-2k)/(k+10)`

Similarly multiplying (A) by `k` and (B) by `5` and adding two together, we get

`(3k+30)secalpha=k^2+25` and `secalpha=(k^2+25)/(3k+30)`

Now as `sec^2alpha=tan^2alpha+1`, we have

`((k^2+25)/(3k+30))^2=((5-2k)/(k+10))^2+1`

or `(k^2+25)^2=9(5-2k)^2+(3k+30)^2`

or `k^4+50k^2+625=225-180k+36k^2+9k^2+180k+900`

or `k^4+5k^2-500=0`

or `(k^2+25)(k^2-20)=0`

As `k^2+25!=0`, we have `k^2=20`

and answer is (a).

Answer 2

`k^2=20`

Explanation:

.

`3secalpha-5tanalpha=k`

`6secalpha+ktanalpha=5`

Multiplying the first equation by `-2`, we get:

`-6secalpha+10tanalpha=-2k`

`6secalpha+ktanalpha=5`

Adding the two equations together, we get:

`(10+k)tanalpha=5-2k`

`tanalpha=(5-2k)/(10+k)`

From the first equation, we get:

`secalpha=(k+5tanalpha)/3`

Let's plug in the value of `tanalpha`:

`secalpha=(k+5((5-2k)/(10+k)))/3=(k^2+25)/(3(10+k))`

`1/cosalpha=(k^2+25)/(3(10+k))`

`cosalpha=(3(10+k))/(k^2+25)`

`cos^2alpha=(9(10+k)^2)/(k^2+25)^2`

Since `tanalpha=sinalpha/cosalpha`, we can say:

`(5-2k)/(10+k)=sinalpha/((3(10+k))/(k^2+25)`

`(5-2k)/(10+k)=sinalpha((k^2+25)/(3(10+k)))`

Let's multiply both sides by `3(10+k)`

`3(5-2k)=sinalpha(k^2+25)`

`sinalpha=(3(5-2k))/(k^2+25)`

`sin^2alpha=(9(5-2k)^2)/(k^2+25)^2`

Using identity `sin^2alpha+cos^2alpha=1`, we can substitute values:

`(9(5-2k)^2)/(k^2+25)^2+(9(10+k)^2)/(k^2+25)^2=1`

Multiplying both sides by `(k^2+25)^2`, we get:

`9(5-2k)^2+9(10+k)^2=(k^2+25)^2`

Simplifying, we get:

`45(k^2+25)=(k^2+25)^2`

Dividing both sides by `(k^2+25)`, we get:

`45=k^2+25`

`k^2=20`