Chỉ mình câu c, d với! Chứng minh các đẳng thức

Đáp án:

Giải thích các bước giải:1
−
tan
a
tan
b
=
1
−
sin
a
cos
a
.
sin
b
cos
b
=
cos
a
.
cos
b
−
sin
a
.
sin
b
cos
a
cos
b
=
cos
(
a
+
b
)
cos
a
cos
b
1−tan⁡atan⁡b=1−sin⁡acos⁡a.sin⁡bcos⁡b=cos⁡a.cos⁡b−sin⁡a.sin⁡bcos⁡acos⁡b=cos⁡(a+b)cos⁡acos⁡b
1
+
tan
a
tan
b
=
1
+
sin
a
cos
a
.
sin
b
cos
b
=
cos
a
.
cos
b
+
sin
a
.
sin
b
cos
a
cos
b
=
cos
(
a
−
b
)
cos
a
cos
b
1+tan⁡atan⁡b=1+sin⁡acos⁡a.sin⁡bcos⁡b=cos⁡a.cos⁡b+sin⁡a.sin⁡bcos⁡acos⁡b=cos⁡(a−b)cos⁡acos⁡b
→
(
1
−
tan
a
tan
b
)
(
1
+
tan
a
tan
b
)
=
cos
(
a
+
b
)
cos
a
cos
b
.
cos
(
a
−
b
)
cos
a
cos
b
→(1−tan⁡atan⁡b)(1+tan⁡atan⁡b)=cos⁡(a+b)cos⁡acos⁡b.cos⁡(a−b)cos⁡acos⁡b
1
−
tan
2
a
tan
2
b
=
cos
(
a
+
b
)
cos
(
a
−
b
)
cos
2
a
cos
2
b
1−tan2⁡atan2⁡b=cos⁡(a+b)cos⁡(a−b)cos2⁡acos2⁡b
d.Ta có :
tan
a
+
tan
b
=
sin
a
cos
a
+
sin
b
cos
b
=
sin
a
cos
b
+
sin
b
cos
a
cos
a
cos
b
=
sin
(
a
+
b
)
cos
a
cos
b
tan⁡a+tan⁡b=sin⁡acos⁡a+sin⁡bcos⁡b=sin⁡acos⁡b+sin⁡bcos⁡acos⁡acos⁡b=sin⁡(a+b)cos⁡acos⁡b
tan
a
−
tan
b
=
sin
a
cos
a
−
sin
b
cos
b
=
sin
a
cos
b
−
sin
b
cos
a
cos
a
cos
b
=
sin
(
a
−
b
)
cos
a
cos
b
tan⁡a−tan⁡b=sin⁡acos⁡a−sin⁡bcos⁡b=sin⁡acos⁡b−sin⁡bcos⁡acos⁡acos⁡b=sin⁡(a−b)cos⁡acos⁡b
→
(
tan
a
+
tan
b
)
(
tan
a
−
tan
b
)
=
sin
(
a
−
b
)
sin
(
a
+
b
)
cos
2
a
cos
2
b
→(tan⁡a+tan⁡b)(tan⁡a−tan⁡b)=sin⁡(a−b)sin⁡(a+b)cos2⁡acos2⁡b
→
tan
2
a
−
tan
2
b
=
sin
(
a
−
b
)
sin
(
a
+
b
)
cos
2
a
cos
2
b
→tan2⁡a−tan2⁡b=sin⁡(a−b)sin⁡(a+b)cos2⁡acos2⁡b
→
đ
p
c
m