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51.

If `tan theta = (sin alpha - cos alpha) /(sin alpha + cos alpha)`, then `sin alpha+cos alpha` is

Answer» `tan theta = ( sin alpha/cos alpha - 1)/(sin alpha/cos alpha + 1)`
`tan theta = ( tan alpha - 1)/(tan alpha + 1)`
`tan theta = ( tan alpha - tan (pi/4))/( 1 + tan alpha* tan(pi/4))`
`tan theta = tan(alpha - pi/4)`
`theta = alpha - pi/4`
`alpha = theta + pi/4`
now,`sin(theta+ pi/4) + cos(theta + pi/4)`
`= sin theta 1/sqrt2 + cos theta 1/sqrt2 + 1/sqrt2 cos theta - 1/sqrt2 sin theta`
`= 2*1/sqrt2 cos theta`
`= sqrt2 cos theta`
option 2 is correct
Discussion
52.

The minimum value of `cos^2 theta+ sec^2 theta` is

Answer» We know, arithmatic mean of two terms is always greater than or equal to their geometric mean.
`:. 1/2(cos^2theta+sec^2theta) ge (cos^2thetasec^2theta)^(1/2)`
`=> (cos^2theta+sec^2theta) ge 2(cos^2theta*1/cos^2theta)^(1/2)`
`=> (cos^2theta+sec^2theta) ge 2`
So, minimum value of `cos^2theta+sec^2theta` is `2`.
Discussion
53.

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.(x) `((1+tan^2A)/(1+cot^2A))=((1-tanA)/(1-cotA))^2=tan^2A`

Answer» Right hand side
=`((1+tan^2A)/(1+cot^2A))`
=`((1+tan^2A)/((tan^2A+1)/tan^2A))`
=`tan^2A`
Left hand side
=`((1-tan^2A)/(1-cot^2A))`
=`((1-tan^2A)/((tan^2A-1)/tan^2A))`
=`tan^2A`
Discussion
54.

If tanA=ntanB and sinA=msinB, prove that `cos^2A=[m^2-1]/[n^2-1]`

Answer» `tanA = ntanB=> n = tanA/tanB`
`sinA = msinB => m = sinA/sinB`
`R.H.S. = (m^2-1)/(n^2-1) = (sin^2A/sin^2B-1)/(tan^2A/tan^2B-1)`
`=((sin^2A-sin^2B)/sin^2B)/((tan^2A-tan^2B)/tan^2B)`
`=((sin^2A-sin^2B)/sin^2B)/((tan^2A-tan^2B)/(sin^2B/cos^2B))`
`=(sin^2A-sin^2B)/(cos^2B(tan^2A-tan^2B))`
`=(1-cos^2A-(1-cos^2B))/(cos^2B(sec^2A-1-(sec^2B-1)))`
`=(cos^2B-cos^2A)/(cos^2B(sec^2A-sec^2B))`
`=(cos^2B-cos^2A)/(cos^2B(1/cos^2A-1/cos^2B)`
`=(cos^2Acos^2B(cos^2B-cos^2A))/(cos^2B(cos^2B-cos^2A)`
`=cos^2A = L.H.S.`
Discussion
55.

If `cosectheta + cottheta=p`, then prove that the `cos theta=(p^2-1)/(p^2+1)`

Answer» Given, `cosectheta + cottheta=p`
`rArr 1/(sintheta)+ (costheta)/(sintheta)=p` `[therefore cosectheta=1/(sintheta)` and `cottheta=(costheta)/(sintheta)]`
`rArr (1+costheta)/(sintheta)=p/1`
`rArr (1+costheta)^(2)/(sin^(2)theta)=p^(2)/1` [take square on both sides]
`rArr (1+ cos^(2)theta + 2 cos theta)/(sin^(2)theta)=p^(2)/1`
Using componendo and dividendo rule, we get
`((1+cos^(theta)+2costheta)-sin^(2)theta)/((1+cos^(2)theta+2costheta)+sin^(2)theta)= (p^(2)-1)/(p^(2)+1)`
`rArr (1+cos^(2)theta + 2costheta-(1-cos^(2)theta))/(1+2costheta+(cos^(2)theta+sin^(2)theta))=(p^(2)-1)/(p^(2)+1)` `[therefore sin^(2)theta + cos^(2)theta=1]`
`rArr (2cos^(2)theta + 2costheta)/(2+2costheta) = (p^(2)-1)/(p^(2)+1)`
`rArr (2costheta(costheta+1))/(2(costheta+1))=(p^(2)-1)/(p^(2)+1)`
`therefore costheta=(p^(2)-1)/(p^(2)+1)` Hence proved.
Discussion
56.

State whether the following are true or false. Justify your answer. (i) The value of tan A always less than 1.(ii) `sec A=(12)/5`for some value of angle A(iii) cos A is the abbreviation used for the cosecant of angle A.(iv) cot A is the product of cot and A(v)`sin theta = 4/3` for some angle `theta`

Answer» 1) False, the value of tan A can be greater than 1.
2)True,
3)True,
4)False,Cot A is a tangent of A.
5)False,because Sin `theta` lies between-1 to 1.
Discussion
57.

In a right triangle ABC right-angled at B. if `t a n A=1`, then verify that `2sinA cos A=1.`

Answer» `tanA=(BC)/(AB)`
`1=(BC)/(AB)`
`BC=AB`
`AC=sqrt(AB^2+BC^2)`
`=ABsqrt2`
`=sqrt3BC`
`SinA=(BC)/(AC)=(BC)/(sqrt2BC)=1/sqrt2
``CosA=(AB)/(AC)=(AB)/(sqrt2BC)=1/sqrt2`
`2CosASinA=2*1/sqrt2*1/sqrt2=1`
Discussion
58.

In `DeltaP Q R ,`right-angled at Q (see figure), `P Q = 3 c m a n d P R = 6 c m`. Determine `/_Q P R`and `/_P R Q`

Answer» If we create a right angle triangle `Delta PQR` with given details,
then,
`sinR = PQ/PR = 3/6 = 1/2 =sin30^@=> R =30^@`
We know, `/_P+/_Q+/_R = 180^@`
`/_P+90+30 = 180=>/_P = 180-120=60^@`
So,`/_QPR = 60^@ and /_PRQ = 30^@`
Discussion