find the value of tan teeta _ cot (90degree theeta)
plz answer it question guys fast
Answers
Answer:
hi mate here is you answer
Explanation:
sin12cos15sec78cosec75)
[cot(90−θ)tanθ−cosec(90−θ)secθ]
+
tan15tan37tan53tan75
cos
2
(50+θ)tan
2
(40−θ)
=−1+sin
2
(40−θ)tan
2
(40−θ)
Step-by-step explanation:
\begin{gathered} Value \: of \: \\\frac{[cot(90-\theta)tan\theta-cosec(90-\theta)sec\theta]}{sin12 cos15 sec78 cosec75)}+\frac{cos^{2}(50+\theta)tan^{2}(40-\theta)}{tan15 tan37 tan53 tan75}\end{gathered}
Valueof
sin12cos15sec78cosec75)
[cot(90−θ)tanθ−cosec(90−θ)secθ]
+
tan15tan37tan53tan75
cos
2
(50+θ)tan
2
(40−θ)
=\frac{tan\theta tan\theta-sec\theta sec\theta}{sin12sec(90-12)cos15cosec(90-15)}+\frac{cos^2[90-(40-\theta)]tan^{2}(40-\theta)}{tan15tan(90-15)tan37tan(90-37)}=
sin12sec(90−12)cos15cosec(90−15)
tanθtanθ−secθsecθ
+
tan15tan(90−15)tan37tan(90−37)
cos
2
[90−(40−θ)]tan
2
(40−θ)
=\frac{(tan^{2}\theta -sec^{2}\theta)}{(sin12cosec12)(cos15sec15)}+\frac{sin^2(40-\theta)tan^{2}(40-\theta)}{(tan15cot15)(tan37cot37)}=
(sin12cosec12)(cos15sec15)
(tan
2
θ−sec
2
θ)
+
(tan15cot15)(tan37cot37)
sin
2
(40−θ)tan
2
(40−θ)
=\frac{-1}{1 \times 1}+\frac{sin^2(40-\theta)tan^{2}(40-\theta)}{1 \times 1}=
1×1
−1
+
1×1
sin
2
(40−θ)tan
2
(40−θ)
=-1+sin^2(40-\theta)tan^{2}(40-\theta)=−1+sin
2
(40−θ)tan
2
(40−θ)
Therefore,
\begin{gathered} Value \: of \: \\\frac{[cot(90-\theta)tan\theta-cosec(90-\theta)sec\theta]}{sin12 cos15 sec78 cosec75)}+\frac{cos^{2}(50+\theta)tan^{2}(40-\theta)}{tan15 tan37 tan53 tan75} \\\\=-1+sin^2(40-\theta)tan^{2}(40-\theta)\end{gathered}
Valueof
sin12cos15sec78cosec75)
[cot(90−θ)tanθ−cosec(90−θ)secθ]
+
tan15tan37tan53tan75
cos
2
(50+θ)tan
2
(40−θ)
=−1+sin
2
(40−θ)tan
2
(40−θ)
•••♪
Answer:
0
Explanation:
tan theta -cot 90-theta
=tan theta- tan theta-
=0