问题11。如果secsec A + sec A = secsec B + sec B,则证明tan A tan B = 。
解决方案:
We have, cosec A + sec A = cosec B + sec B
=> sec A − sec B = cosec B − cosec A
=>
=>
=>
=> tan A tan B =
=> tan A tan B =
Hence proved.
问题12.如果sin 2A =λsin 2B,则证明 。
解决方案:
We are given, sin 2A = λ sin 2B
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=>
Hence proved.
问题13:证明:
(一世)
解决方案:
We have,
L.H.S. =
=
=
=
=
=
=
=
= cot C
= R.H.S.
Hence proved.
(ii)sin(BC)cos(AD)+ sin(CA)cos(BD)+ sin(AB)cos(CD)= 0
解决方案:
We have, L.H.S. = sin (B−C) cos (A−D) + sin (C−A) cos (B−D) + sin (A−B) cos (C−D)
=
=
=
=
= 0
= R.H.S.
Hence proved.
问题14.如果 ,证明tan A tan B tan C tan D = -1。
解决方案:
We have,
=>
=>
=>
=>
=>
=>
=>
=>. . . . (1)
Also,
=>
=>
=>
=>
=>. . . . (2)
Dividing (1) by (2), we get,
=>
=>
=>
=> tan A tan B tan C tan D = −1
Hence proved.
问题15.如果cos( α+β)sin(γ+δ)= cos(α-β)sin(γ-δ),则证明cotαcotβcotγ= cotδ。
解决方案:
We have, cos (α+β) sin(γ+δ) = cos (α−β) sin(γ−δ)
=>
=>
=>
=>
=>. . . . (1)
Also,
=>
=>
=>
=>. . . . (2)
Dividing (1) by (2), we get,
=>
=>
=> cot α cot β = tan γ cot δ
=> cot α cot β cot γ = cot δ
Hence proved.
问题16。如果y sinØ= x sin(2θ +Ø),则证明(x + y)cot( θ +Ø)=(y − x)cotθ 。
解决方案:
Given, y sin Ø = x sin (2θ + Ø)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=> tan (Ø+θ) cot θ =
=>
=> (y − x) cot θ = (x + y) cot (θ + Ø)
=> (x + y) cot (θ + Ø) = (y − x) cot θ
Hence proved.
问题17.如果cos(A + B)sin(C − D)= cos(AB)sin(C + D),则证明tan A tan B tan C + tan D = 0。
解决方案:
We are given, cos (A+B) sin (C−D) = cos (A−B) sin (C+D)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=> −tan D = tan A tan B tan C
=> tan A tan B tan C + tan D = 0
Hence proved.
问题18 ,证明xy + yz + zx = 0。
解决方案:
We have,= k (say)
x =
y =
z =
So, L.H.S. = xy + yz + zx
=
=
=
=
=
=
= 0
= R.H.S.
Hence proved.
问题19.如果m sinθ = n sin( θ + 2a),则证明 。
解决方案:
We are given, m sin θ = n sin (θ + 2a)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=>
Hence, proved.